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\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
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 .............
a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
a) điều kiện xác định \(x-2\ge0vàx^2-4x+3\ge0\)
\(pt\Leftrightarrow x^2-4x+3=x-2\Leftrightarrow x^2-5x+5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{5}}{2}\\x=\dfrac{5-\sqrt{5}}{2}\left(L\right)\end{matrix}\right.\) bạn giải nó bằng cách giải den ta nha .
vậy \(x=\dfrac{5+\sqrt{5}}{2}\)
b) điều kiện xác định : \(x\ge1\)
đặc \(\sqrt{x-1}=t\left(t\ge0\right)\)
\(pt\Leftrightarrow2\left(\dfrac{t}{2}-3\right)=\dfrac{2.2t}{3}-\dfrac{1}{3}\) giải phương trình này rồi thế ngược lại là xong
c) điều kiện xác định : \(x\ge\dfrac{7}{9}\)
\(pt\Leftrightarrow9x-7=7x+5\Leftrightarrow x=6\) vậy \(x=6\)
d) câu cuối chờ nhát h mk chưa nghỉ ra
d) Ta có pt \(4+\sqrt{2x+6-6\sqrt{2x-3}}=\sqrt{2x-2+2\sqrt{2x-3}}=0\)
\(\Leftrightarrow4+\sqrt{2x-3-6\sqrt{2x-3}+9}=\sqrt{2x-3-2\sqrt{2x-3}+1}\Leftrightarrow4+\left|\sqrt{2x-3}-3\right|=\left|\sqrt{2x-3}-1\right|\)
Đặt \(\sqrt{2x-3}=a\left(a\ge0\right),pt\Leftrightarrow4+\left|a-3\right|=\left|a-1\right|\)
xét \(a\ge3,pt\Leftrightarrow4+a-3=a-1\Leftrightarrow0a=1\left(VN\right)\)
xét \(a\le1.pt\Leftrightarrow4+3-a=1-a\Leftrightarrow0a=6\left(VN\right)\)
xét \(3>x>1,pt\Leftrightarrow4+3-a=a-1\Leftrightarrow a=1\)(k thỏa mãn )
=> pt vô nghiệm !
1,
a,\(4\sqrt{\dfrac{9}{2}}+\sqrt{2}+\sqrt{\dfrac{1}{18}}=4\sqrt{\dfrac{18}{4}}+\sqrt{2}+\sqrt{\dfrac{1}{9.2}}=4\dfrac{\sqrt{18}}{2}+\sqrt{2}+\dfrac{1}{3}\sqrt{\dfrac{1}{2}}=2\sqrt{9.2}+\sqrt{2}+\dfrac{1}{3}\sqrt{\dfrac{2}{4}}=2.3\sqrt{2}+\sqrt{2}+\dfrac{\sqrt{2}}{6}=6\sqrt{2}+\sqrt{2}+\sqrt{2}\dfrac{1}{6}=\dfrac{43}{6}\sqrt{2}\) b,\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}=4\sqrt{4.5}-3\sqrt{25.5}+5\sqrt{9.5}-15\dfrac{\sqrt{5}}{5}=4.2\sqrt{5}-3.5\sqrt{5}+5.3\sqrt{5}-3\sqrt{5}=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
*) Giải phương trình :
\(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\) ( ĐKXĐ : x \(\ge\) 2 )
\(\Leftrightarrow\sqrt{4\left(x-2\right)}+5\sqrt{x-2}-\sqrt{9\left(x-2\right)}=20\)
\(\Leftrightarrow2\sqrt{x-2}+5\sqrt{x-2}-3\sqrt{x-2}=20\)
\(\Leftrightarrow4\sqrt{x-2}=20\)
\(\Leftrightarrow\sqrt{x-2}=5\)
\(\Leftrightarrow x-2=25\)
\(\Leftrightarrow x=27\) ( thỏa mãn điều kiện )
Vậy phương trình có nghiệm x = 27 .
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
a.\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\\ \sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\\ 3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\\ 2\sqrt{x+2}=6\\ \left\{{}\begin{matrix}x\ge2\\x+2=36\end{matrix}\right.\\ \left\{{}\begin{matrix}x>=2\\x=34\end{matrix}\right.\\ \)
Vậy.....
\(2,\\ a,PT\Leftrightarrow\sqrt{\left(5x-1\right)^2}=\sqrt{4\left(x+1\right)^2}\\ \Leftrightarrow\left|5x-1\right|=2\left|x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=2\left(x+1\right)\\1-5x=2\left(x+1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{7}\end{matrix}\right.\)
\(b,ĐK:x^2-3\ge0\\ PT\Leftrightarrow\sqrt{x^2-3}=x-1\\ \Leftrightarrow x^2-3=x^2-2x+1\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ c,ĐK:x\le\dfrac{7}{2}\\ PT\Leftrightarrow7-2x=x^2+7\\ \Leftrightarrow x^2+2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ge3\\ PT\Leftrightarrow3\sqrt{x-3}+\dfrac{1}{2}\cdot2\sqrt{x-3}-9\cdot\dfrac{1}{3}\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
thêm bài ở trên mình gửi là xong