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a, ĐKXĐ: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-3}\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\\x=2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
b, ĐKXĐ: \(x\ge-1\)
\(pt\Leftrightarrow\sqrt{x+1}\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\x+1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ: \(x>2\)
\(pt\Leftrightarrow\frac{x}{\sqrt{x-2}}=\frac{3-x}{\sqrt{x-2}}\)
\(\Leftrightarrow x=3-x\)
\(\Leftrightarrow x=\frac{3}{2}\left(l\right)\)
\(\Rightarrow\) Phương trình vô số nghiệm
d, ĐKXĐ: \(x>-1\)
\(pt\Leftrightarrow\frac{x^2-4}{\sqrt{x+1}}=\frac{x+3+x+1}{\sqrt{x+1}}\)
\(\Leftrightarrow x^2-4=2x+4\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=4\)
Câu a:
ĐKXĐ: \(x\neq \pm 3\)
\(\left|\frac{x+5}{-x^2+9}\right|=2\Rightarrow \left[\begin{matrix} \frac{x+5}{-x^2+9}=2\\ \frac{x+5}{-x^2+9}=-2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x+5=2(-x^2+9)\\ x+5=-2(-x^2+9)\end{matrix}\right.\Rightarrow \left[\begin{matrix} 2x^2+x-13=0\\ 2x^2-x-23=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1\pm \sqrt{105}}{4}\\ x=\frac{1\pm \sqrt{185}}{4}\end{matrix}\right.\) (đều thỏa mãn )
Vậy.......
Câu b:
ĐKXĐ: \(x< 2\)
Ta có: \(\frac{4}{\sqrt{2-x}}-\sqrt{2-x}=2\)
\(\Rightarrow 4-(2-x)=2\sqrt{2-x}\)
\(\Leftrightarrow 4=(2-x)+2\sqrt{2-x}\)
\(\Leftrightarrow 5=(2-x)+2\sqrt{2-x}+1=(\sqrt{2-x}+1)^2\)
\(\Rightarrow \sqrt{2-x}+1=\sqrt{5}\) (do \(\sqrt{2-x}+1>0\) )
\(\Rightarrow \sqrt{2-x}=\sqrt{5}-1\)
\(\Rightarrow 2-x=6-2\sqrt{5}\)
\(\Rightarrow x=-4+2\sqrt{5}\) (thỏa mãn)
Vậy...........
a. R / \(\left\{-2\right\}\)
b. R / \(\left\{4;-1\right\}\)
c. R ( mẫu luôn > 0 )
d. \(\left(2;+\infty\right)\)
e. \(\left(-\infty;\dfrac{5}{6}\right)\)
f. \(\left(2;+\infty\right)\)
g. \(\left(1;3\right)\)
h. \(\left(5;+\infty\right)\)
i. \(\left(1;+\infty\right)\)
k. \(\left(-\infty;2\right)\)
l. R/\(\left\{\pm3\right\}\)
m. \(\left(-2;+\infty\right)/\left\{3\right\}\)
a/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
Ta được:
\(1+\frac{a^2-1}{3}=a\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x\left(1-x\right)}=0\\2\sqrt{x-x^2}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\left(1-x\right)=0\\-4x^2+4x-9=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\sqrt{x+5}=a\ge0\Rightarrow a^2-x=5\)
\(x^2+a=a^2-x\)
\(\Leftrightarrow x^2-a^2+a+x=0\)
\(\Leftrightarrow\left(a+x\right)\left(x-a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-x\left(x\le0\right)\\\sqrt{x+5}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2\left(x\le0\right)\\x+5=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
c/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
b: ĐKXĐ: x>=-1
\(\sqrt{x+1}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+1\right)^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\cdot x=0\\x>=-1\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1\right\}\)
c: \(\sqrt{x-1}=1-x\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\1-x< =0\end{matrix}\right.\Leftrightarrow x=1\)
Do đó: x=1 là nghiệm của phương trình
d: \(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)(ĐKXĐ: x<>1)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)
\(\Leftrightarrow2x^2-2x+3x-3+4-x^2-3=0\)
\(\Leftrightarrow x^2+x-2=0\)
=>(x+2)(x-1)=0
=>x=-2(nhận) hoặc x=1(loại)