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a) \(\sqrt{4x}=10\) (ĐKXĐ: 4x>=0 <=> x>=0)
\(\Leftrightarrow4x=100\)
\(\Leftrightarrow x=25\)
\(S=\left\{25\right\}\)
b) \(\sqrt{x^2-2x+1}=8\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=8\)
\(\Leftrightarrow x-1=8\)
\(\Leftrightarrow x=9\)
\(S=\left\{9\right\}\)
c) \(\sqrt{x^2-6x+9}=\sqrt{1-6x+9x^2}\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(1-3x\right)^2}\)
\(\Leftrightarrow x-3=1-3x\) hoặc \(\Leftrightarrow x-3=-1+3x\)
\(\Leftrightarrow x+3x=1+3\) \(\Leftrightarrow x-3x=-1+3\)
\(\Leftrightarrow4x=4\) \(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=1\) \(\Leftrightarrow x=-1\)
\(S=\left\{1;-1\right\}\)
d) \(\sqrt{2x-5}=x-2\)
\(\Leftrightarrow2x-5=x^2-4x+4\)
\(\Leftrightarrow-x^2+2x+4x-5-4=0\)
\(\Leftrightarrow-x^2+6x-9=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
e) \(\sqrt{x^2-2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x+1=x+1\)
\(\Leftrightarrow x^2-2x-x+1-1=0\)
\(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{0;3\right\}\)
g) \(\sqrt{x^2-9}-\sqrt{x-3}=0\) ( ĐKXĐ: x-3>=0 <=> x>=3)
\(\Leftrightarrow\sqrt{x^2-9}=\sqrt{x-3}\)
\(\Leftrightarrow x^2-9=x-3\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x+2=0\) hoặc \(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=-2\) \(\Leftrightarrow x=3\)
\(S=\left\{-2;3\right\}\)
h) \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow x-2+x-3-1=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
i) \(\sqrt{\frac{2x-3}{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow4\left(x-1\right)=2x-3\)
\(\Leftrightarrow4x-4-2x+3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(S=\left\{\frac{1}{2}\right\}\)
l) \(x+y+12=4\sqrt{x}+6\sqrt{y-1}\)
\(\Leftrightarrow x+y-4\sqrt{x}+12-6\sqrt{y-1}=0\)
\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left(y-1-6\sqrt{y-1}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\) hoặc \(\Leftrightarrow\sqrt{y-1}-3=0\)
\(\Leftrightarrow\sqrt{x}=2\) \(\Leftrightarrow\sqrt{y-1}=3\)
\(\Leftrightarrow x=4\) \(\Leftrightarrow y-1=9\)
\(\Leftrightarrow y=10\)
KẾT luận : ..............
Tới đây nhé, nếu mai chưa ai giải thì mình giải hộ cho
CHÚC BẠN HỌC TỐT!
m) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
<=> \(\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)+6\sqrt{x-1}+9}=5\)
<=>\(\sqrt{\left(\sqrt{x-1}+2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
<=>\(\sqrt{x-1}+2+\sqrt{x-1}+3=5\)
<=> \(2\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}=0\) <=>x=1
Vậy \(S=\left\{1\right\}\)
n) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\) (*) ( đk \(x\ge\frac{1}{2}\))
<=> \(\left(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}\right)^2=2\)
<=> \(x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{x^2-2x+1}=2\)
<=> 2x+\(2\sqrt{\left(x-1\right)^2=2}\)
<=> x+\(\left|x-1\right|=2\)(1)
TH1: \(\frac{1}{2}\le x\le1\)
Từ (1) => x+1-x=2
<=> 1=2(vô lý)
TH2: x>1
Từ (1)=> x+x-1=2
<=> 2x=3<=> \(x=\frac{2}{3}\)(tm pt (*))
Vậy \(S=\left\{\frac{2}{3}\right\}\)
p) \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\) (*) (đk :\(x\ge2\))
Đặt \(\left\{{}\begin{matrix}x-2=a\left(a\ge0\right)\\x+1=b\left(b\ge0\right)\end{matrix}\right.\) =>a+b=2x-1
Có \(\sqrt{a+b}+\sqrt{a}=\sqrt{b}\)
<=> \(\sqrt{a+b}=\sqrt{b}-\sqrt{a}\)
<=> \(a+b=b-2\sqrt{ab}+a\)
<=> 0=\(-2\sqrt{ab}\)
=> \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) => x=2 (vì x=-1 không thỏa mãn pt(*))
Vậy \(S=\left\{2\right\}\)
q) \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)(*) (đk : \(7\le x\le9\))
Với a,b\(\ge0\) có: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\)(tự cm nha) .Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên có:
\(\sqrt{x-7}+\sqrt{9-x}\le2\sqrt{\frac{x-7+9-x}{2}}=2\sqrt{\frac{2}{2}}=2\) (1)
Có x2-16x+66=(x2-16x+64)+2=(x-8)2+2 \(\ge2\) với mọi x (2)
Từ (1),(2) .Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-7=9-x\\x-8=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}2x=16\\x=8\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=8\\x=8\end{matrix}\right.\)<=> x=8( tm pt (*))
Vậy \(S=\left\{8\right\}\)
\(a,\sqrt{x-2\sqrt{x}-1}-\sqrt{x-1}=1.\)
\(\Rightarrow\sqrt{\left(\sqrt{x}-1\right)^2}-\sqrt{x-1}=1\)
\(\Rightarrow x-1-\sqrt{x-1}=1\)
\(\Rightarrow\sqrt{x-1}=x-1+1\)
\(\Rightarrow x-1=x^2\Rightarrow x^2-x+1=0\) ( vô nghiệm vì nó luôn lớn hơn 0 )
\(đkxđ\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)
\(c,\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}.\)
\(\Rightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Rightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Rightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Rightarrow\sqrt{2x-1}+1+\sqrt{2x-1}-1=2\)
\(\Rightarrow\sqrt{2x-1}+\sqrt{2x-1}=2\)
\(\Rightarrow\sqrt{2x-1}=1\Rightarrow\sqrt{2x-1}^2=1\)
\(\Rightarrow2x-1=1\Rightarrow2x=2\Leftrightarrow x=1\)\(\left(tm\right)\)
d tương tự nha , nhân thêm 2 vế với \(\sqrt{6}\)là ra
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
Làm hơi tắt xíu, có gì ko hiểu cmt nha :>
\(a.\sqrt{x-1}=3\left(ĐK:x\ge1\right)\Leftrightarrow x-1=9\Leftrightarrow x=10\)
\(b.\sqrt{x^2-4x+4}=2\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=2\\ \Leftrightarrow\left|x-2\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2\left(x\ge2\right)\\2-x=2\left(x< 2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
\(c.\sqrt{25x^2-10x+1}=4x-9\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x-9\\ \Leftrightarrow\left|5x-1\right|=4x-9\\\Leftrightarrow \left[{}\begin{matrix}5x-1=4x-9\left(x\ge\frac{1}{5}\right)\\1-5x=4x-9\left(x< \frac{1}{5}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\left(ktm\right)\\x=\frac{10}{9}\left(ktm\right)\end{matrix}\right.\)
\(d.\sqrt{x^2+2x+1}=\sqrt{x+1}\left(ĐK:x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e. ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\)
Câu cuối chưa nghĩ ra, sorry :<
a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)
\(=\sqrt{1+\sqrt{2}+\sqrt{3}-\left(\sqrt{3}+\sqrt{2}\right)}=1\)
b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)
\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)
Th1: x-3 < 0
\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)
Th2: x-3 > 0
\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)
c)
Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\dfrac{\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
Th1: \(x-2\ge0\Leftrightarrow x\ge2\)
\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)
Th2: \(x-2\le0\Leftrightarrow x\le2\)
kết hợp với đk, ta được: 1 \< x \< 2
\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)
d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
chẳng biết có sai sót gì 0 nữa, xin lỗi tớ 0 xem lại đâu vì chán quá!
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\) \(+,x\ge2\Rightarrow\left\{{}\begin{matrix}x-2\ge0\\x-1\ge1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|=x-2\\\left|x-1\right|=x-1\end{matrix}\right.\Rightarrow\left|x-2\right|+\left|x-1\right|=x-2+x-1=3\Leftrightarrow2x-3=3\Leftrightarrow x=3\left(\text{t/m}\right)\) \(+,1\le x< 2\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=x-1+2-x=1\left(l\right)\) \(+,x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=-\left(x-1\right)=1-x\\\left|x-2\right|=-\left(x-2\right)=2-x\end{matrix}\right.\Rightarrow\left|x-1\right|+\left|x-2\right|=1-x+2-x=3\Leftrightarrow3-2x=3\Leftrightarrow x=0\left(\text{t/m}\right)\) \(f,\left\{{}\begin{matrix}\sqrt{x^2-9}\ge0\\\sqrt{x^2-6x+9}\ge0\end{matrix}\right.mà:\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-9}=0\\\sqrt{x^2-6x+9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\sqrt{\left(x-3\right)^2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-9=0\\\left|x-3\right|=0\end{matrix}\right.\Leftrightarrow x=3\)\thay vào ta thấy thoa man => x=3
\(ĐK:x\ge4\)\(\sqrt{x^2+x-20}=\sqrt{x^2+5x-4x-20}=\sqrt{x\left(x+5\right)-4\left(x+5\right)}=\sqrt{\left(x-4\right)\left(x+5\right)}=\sqrt{x-4}.\sqrt{x+5}=\sqrt{x-4}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}=0\\\sqrt{x+5}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x+5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-4\left(l\right)\end{matrix}\right.\Rightarrow x=4\) \(b,ĐK:x\le2;\sqrt{x+1}+\sqrt{2-x}=\sqrt{6}\Leftrightarrow x+1+2-x+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow3+2\sqrt{\left(x+1\right)\left(2-x\right)}=6\Leftrightarrow2\sqrt{\left(x+1\right)\left(2-x\right)}=3\Leftrightarrow\sqrt{\left(x-1\right)\left(2-x\right)}=1,5\Leftrightarrow\left(x-1\right)\left(2-x\right)=\frac{9}{4}\Leftrightarrow\left(x-1\right)\left(x-2\right)=-\frac{9}{4}\Leftrightarrow x^2-3x+2=-\frac{9}{4}\Leftrightarrow x^2-3x+\frac{9}{4}=-2\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-2\Rightarrow vonghiem\)
a,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) (*)(đk \(x\ge-2\))
<=> \(\sqrt{\left(x+2\right)-4\sqrt{x+2}+4}+\sqrt{\left(x+2\right)-6\sqrt{x+2}+9}\)=1
<=> \(\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|\)=1 (1)
TH1: \(0\le\sqrt{x+2}< 2\)
Từ (1) =>\(2-\sqrt{x+2}+3-\sqrt{x+2}=1\)
<=> \(5-2\sqrt{x+2}=1\) <=> \(2\sqrt{x+1}=4\) <=> \(\sqrt{x+1}=2\)
<=> \(x+1=4\) <=> x=3(không t/m \(\sqrt{x+2}\le2\))
TH2 : \(2\le\sqrt{x+2}\le3\)
Từ (1) =>\(\sqrt{x+2}-2+3-\sqrt{x+2}=1\)
<=> \(1=1\) (luôn đúng)
Từ TH2 <=> 4\(\le x+2\le9\) <=> \(2\le x\le7\)
TH3 \(\sqrt{x+2}>3\)
Từ (1) => \(\sqrt{x+2}-2+\sqrt{x+2}-3=1\)
<=> \(2\sqrt{x+2}=6\) <=> \(\sqrt{x+2}=3\) <=> \(x+2=9\) <=> x=7 (không t/m \(\sqrt{x+2}>3\))
Vậy pt (*) có tập nghiệm S=\(\left\{2\le x\le7\right\}\)
b, \(x^2-10x+27=\sqrt{6-x}+\sqrt{x-4}\) (*) (đk :\(4\le x\le6\))
Vs a,b \(\ge0\) ta có \(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a^2+b^2\right)}\)(tự CM nha)
Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên ta có: \(\sqrt{6-x}+\sqrt{x-4}\le\sqrt{2\left(6-x+x-4\right)}=\sqrt{2.2}=2\)
<=> \(\sqrt{6-x}+\sqrt{x-4}\le2\)(1)
Lại có: \(x^2-10x+27=x^2-10x+25+2=\left(x-5\right)^2+2\ge2\)
<=> \(x^2-10x+27\ge2\) (2)
Từ (1),(2) => Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}6-x=x-4\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}6+4=2x\\x=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\left(tm\right)\)
Vậy pt (*) có tập nghiệm S=\(\left\{5\right\}\)
c, \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)(*) (đk: x\(\ge0\))
<=> \(x\left(x-2\right)-\sqrt{x}\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\left(x-\sqrt{x}\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\sqrt{x}\left(\sqrt{x}-1\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\left(\sqrt{x}-1\right)\left[\sqrt{x}\left(x-2\right)-4\right]=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}\left(x-2\right)-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}\left(x-2\right)=4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x\left(x-2\right)^2=16\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\x\left(x^2-4x+4\right)-16=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=1\\x^3-4x^2+4x-16=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=1\\x^2\left(x-4\right)+4\left(x-4\right)=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\\left(x^2+4\right)\left(x-4\right)=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\left(tm\right)\)
Vậy pt (*) có tập nghiệm S=\(\left\{1;4\right\}\)
d) x2+3x+1=(x+3)\(\sqrt{x^2+1}\)
<=>(\(\sqrt{x^2+1}-3x+3\sqrt{x^2+1}-\left(x^2+1\right)=0\)
<=>\(\left(\sqrt{x^2+1}-3\right)\left(x-\sqrt{x^2+1}\right)=0\)
<=>\(\sqrt{x^2+1}=3\) hoặc \(x=\sqrt{x^2+1}\)
=>x=\(2\sqrt{2}\)
Giải pt :
1
a. ĐKXĐ : \(x\ge4\)
Ta có :
\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)
\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)
\(\Leftrightarrow x=13\) (TM ĐKXĐ)
Vậy \(S=\left\{13\right\}\)
b.ĐKXĐ : \(-3\le x\le10\)
Ta có :
\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy \(S=\left\{1;6\right\}\)
Câu c,d làm giống câu b
Câu e làm giống câu a