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b) Nhẩm thấy \(x=-2\) là nghiệm, ta xét trường hợp:
* Với \(x>-2\) thì
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}>-1+0+1=0=VP\)
* Với \(x< -2\) thì
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}< -1+0+1=0=VP\)
Do đó pt có nghiệm duy nhất \(x=-2\)
c) Đặt \(\sqrt[4]{1-x}=a;\sqrt[4]{1+x}=b\)
\(\Rightarrow a^4+b^4=2\)
Theo đề bài \(a+b+ab=3\Rightarrow a+b=3-ab\)
Cần giải cái hệ (đợi một xíu em ăn xong em làm tiếp hoặc là nếu bận thì thứ 6 tuần này em làm):v \(\left\{{}\begin{matrix}a^4+b^4=3\\a+b=3-ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a^2+b^2\right)^2=3+2a^2b^2\\ab=3-a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2ab\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2\left(3-a-b\right)\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)
nhìn mà nhác giải vl :v
a) \(\sqrt{3x^2-2x+1}+4x=\sqrt{3x^2+2x}+1\)
<=> \(\sqrt{3x^2-2x+1}=\sqrt{3x^2+2x}+1-4x\)
<=> \(\left(\sqrt{3x^2-2x+1}\right)^2=\left(\sqrt{3x^2+2x}+1-4x\right)^2\)
<=> \(3x^2-2x+1=19x^2-8\sqrt{3x^2+2x}.x-6x+2\sqrt{3x^2+2x}+1\)
<=> \(-16x^2+8\sqrt{3x^2+2x}.x+4x-2\sqrt{3x^2+2x}=0\)
<=> \(-2\left(4x-1\right)\left(2x-\sqrt{3x^2+2x}\right)=0\)
<=> \(\hept{\begin{cases}x=\frac{1}{4}\\x=0\\x=2\end{cases}}\) <=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=0\end{cases}}\) (vì k có ngoặc vuông 3 nên mình dùng tạm ngoặc nhọn, thông cảm)
<=> \(\orbr{\begin{cases}x=\frac{1}{4}\\x=2\end{cases}}\)
b) \(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)
<=> \(\sqrt{x^2+x-2}=\sqrt{2\left(x-1\right)}+1-x^2\)
<=> \(\left(\sqrt{x^2+x-2}\right)^2=\left[\sqrt{2\left(x-1\right)}+1-x^2\right]^2\)
<=> \(x^2+x-2=x^4-2\sqrt{2}.x^2.\sqrt{x-1}-2x^2+2x+2\sqrt{2}.\sqrt{x-2}-1\)
<=> \(x^4-2\sqrt{2}.x^2.\sqrt{x-1}-2x^2+2x+2\sqrt{2}.\sqrt{x-1}-1=x^2+x-2\)
<=> \(-2\sqrt{2}.x^2.\sqrt{x-1}+2\sqrt{2}.\sqrt{x-1}-1=-x^4+3x^2-x-2\)
<=> \(-2\sqrt{2}.x^2.\sqrt{x-1}+2\sqrt{2}.\sqrt{x-1}=-x^4+3x^2-x-1\)
<=> \(-2\sqrt{2}.\sqrt{x-1}.\left(x^2+1\right)=-x^4+3x^2-x-1\)
<=> \(\left[-2\sqrt{2}.\sqrt{x-1}\left(x^2+1\right)\right]^2=\left(-x^4+3x^2-x-1\right)^2\)
<=> \(8x^5-8x^4-16x^3+16x^2+8x-8=x^8-6x^6+2x^5+11x^4-6x^3-5x^2+2x+1\)
<=> x = 1
d) mình làm tắt cho nhanh
d) \(\left(\sqrt{4+x}-1\right)\left(\sqrt{1-x}+1\right)=2x\)
<=> \(\sqrt{4+x}.\sqrt{x-1}+\sqrt{4+x}-\sqrt{x-1}-1=2x\)
<=> \(\sqrt{4+x}.\sqrt{1-x}+\sqrt{4+x}-\sqrt{1-x}=2x+1\)
<=> \(\sqrt{4+x}.\sqrt{x-1}+\sqrt{4+x}=2x+1+\sqrt{x-1}\)
<=> \(\left(\sqrt{4+x}.\sqrt{1-x}+\sqrt{4+x}\right)^2=\left(2x+1+\sqrt{1-x}\right)^2\)
<=> \(2\sqrt{-x+1}.\left(x+4\right)=5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6\)
<=> \(\frac{2\sqrt{-x+1}.\left(x+4\right)}{2\left(x+4\right)}=\frac{5x^2}{2\left(x+4\right)}+\frac{4x\sqrt{-x+1}}{2\left(x+4\right)}+\frac{5x}{2\left(x+4\right)}+\frac{2\sqrt{-2x+1}}{2\left(x+4\right)}-\frac{6}{2\left(x+4\right)}\)
<=> \(\sqrt{-x+1}=\frac{5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6}{2\left(4+x\right)}\)
<=> \(2\sqrt{-x+1}.\left(4+x\right)=5x^2+4x\sqrt{-x+1}+5x+2\sqrt{-x+1}-6\)
<=> \(-2x\sqrt{-x+1}+6\sqrt{-x+1}=5x^2+5x-6\)
<=> \(\frac{2\sqrt{-x+1}.\left(-x+3\right)}{2\left(-x+3\right)}=\frac{5x^2}{2\left(-x+3\right)}+\frac{5x}{2\left(-x+3\right)}-\frac{6}{2\left(-x+3\right)}\)
<=> \(\sqrt{-x+1}=\frac{5x^2+5x-6}{2\left(x-3\right)}\)
<=> \(\left(\sqrt{-x+1}\right)^2=\left[\frac{5x^2+5x-6}{2\left(3-x\right)}\right]^2\)
<=> \(-x+1=\frac{25x^4+50x^3-35x^2-60x+36}{36-24+4x}\)
<=> \(\hept{\begin{cases}x=0\\x=\frac{21}{25}\\x=-3\end{cases}}\)=> x = 21/25 (lý do dùng ngoặc nhọn như lý do mình ghi ở trên =))) )
=> x = 21/25
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
a, \(\sqrt{5+\sqrt{x-1}}\)=6-x
=>bình phương lên => trục \(\sqrt{x-1}\)với x-6 => có nhân tử chung
c, đat \(\sqrt{x^2+7x+7}\)=a => pt 3a2+2a-5=0 => giờ thì đơn giản rồi
b, mk k bít lm
\(\sqrt{x^2-3x+2}-\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
\(\Leftrightarrow\left(\sqrt{x^2-3x+2}-\sqrt{x-2}\right)-\left(\sqrt{x^2+2x-3}+\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)-\left(x-2\right)}{\sqrt{x^2-3x+2}+\sqrt{x-2}}-\dfrac{\left(x^2+2x-3\right)-\left(x+3\right)}{\sqrt{x^2+2x-3}-\sqrt{x+3}}=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x-2}}-\dfrac{\left(x-2\right)\left(x+3\right)}{\sqrt{\left(x+3\right)\left(x-1\right)}-\sqrt{x+3}}=0\)
\(\Leftrightarrow\left(x-2\right)\left[\dfrac{x-2}{\sqrt{x-2}\left(\sqrt{x-1}+1\right)}-\dfrac{x+3}{\sqrt{x+3}\left(\sqrt{x-1}-1\right)}\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\right]=0\)
Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô no
(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))
=> x - 2 = 0
<=> x = 2 (nhận)
\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)
\(\Leftrightarrow\dfrac{\left(4x+1\right)-\left(3x-2\right)}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}\right)\left(x+3\right)=0\)
TH1:
x + 3 = 0
<=> x = - 3 (loại)
TH2:
\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)
\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\)
\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)+\left(\sqrt{3x-2}-2\right)=0\)
\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)
\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{3\left(x-2\right)}{\sqrt{3x-2}+2}=0\)
\(\Leftrightarrow\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}\right)\left(x-2\right)=0\)
Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô no
=> x - 2 = 0
<=> x = 2 (nhận)
~ ~ ~
Vậy x = 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 \(S=\left\{1;6\right\}\)
tth, Hoàng Tử Hà, Bonking, Quoc Tran Anh Le, Vũ Huy Hoàng,
Akai Haruma, @Nguyễn Việt Lâm
giúp mk vs! ngày mai phải nộp r
c) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
\(\Leftrightarrow\sqrt{x^2+6}+2\sqrt{x^2-1}=x\)
\(\Leftrightarrow x^2+6+4\left(x^2-1\right)+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=x^2\)
\(\Leftrightarrow6+4x^2-4+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=0\)
\(\Leftrightarrow4x^2+2+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=0\)
\(\Leftrightarrow2x^2+2\sqrt{\left(x^2+6\right)\left(x^2-1\right)}+1=0\)
Dễ thấy \(VT>0\forall x\)
Do đó pt vô nghiệm
Lời giải:
a)
ĐK: \(0\leq x\leq 1\)
PT \(\Leftrightarrow \sqrt{x+\sqrt{1-x}}=1-\sqrt{x}\)
\(\Rightarrow x+\sqrt{1-x}=1+x-2\sqrt{x}\) (bình phương 2 vế)
\(\Leftrightarrow \sqrt{1-x}-1+2\sqrt{x}=0\)
\(\Leftrightarrow \frac{-x}{\sqrt{1-x}+1}+2\sqrt{x}=0\)
\(\Leftrightarrow \sqrt{x}(2-\frac{\sqrt{x}}{\sqrt{1-x}+1})=0\)
Ta thấy \(\sqrt{1-x}+1\geq 1\Rightarrow \frac{\sqrt{x}}{\sqrt{1-x}+1}\leq \sqrt{x}\leq 1< 2\) với mọi $0\leq x\leq 1$
\(\Rightarrow 2-\frac{\sqrt{x}}{\sqrt{1-x}+1}>0\Rightarrow 2-\frac{\sqrt{x}}{\sqrt{1-x}+1}\neq 0\)
Do đó $\sqrt{x}=0\Leftrightarrow x=0$ là nghiệm duy nhất
b)
ĐK: \(1 \leq x\leq \frac{1+\sqrt{5}}{2}\) hoặc \(0\geq x\geq \frac{1-\sqrt{5}}{2}\)
PT \(\Rightarrow \left\{\begin{matrix} \sqrt{x}-1\geq 0\\ 1-\sqrt{x^2-x}=x-2\sqrt{x}+1\end{matrix}\right.\) (bình phương 2 vế)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1(1)\\ x+\sqrt{x^2-x}-2\sqrt{x}=0(2)\end{matrix}\right.\)
(1) kết hợp với ĐKXĐ suy ra \(1\leq x\leq \frac{1+\sqrt{5}}{2}(*)\)
(2) \(\Leftrightarrow \sqrt{x}(\sqrt{x}+\sqrt{x-1}-2)=0\)
Từ $(*)$ suy ra $x\neq 0$. Do đó \(\sqrt{x}+\sqrt{x-1}-2=0\)
\(\Leftrightarrow \sqrt{x-1}=2-\sqrt{x}\)
\(\Rightarrow x-1=4+x-4\sqrt{x}\) (bình phương)
\(\Leftrightarrow 4\sqrt{x}=5\Rightarrow x=\frac{25}{16}\) (thỏa mãn $(*)$)
Vậy......