Bài 9:
a: ĐKXĐ: \(\begin{cases}5-2x\ge0\\ x-1\ge0\end{cases}\Rightarrow1\le x\le\frac52\)
\(\sqrt{5-2x}=\sqrt{x-1}\)
=>5-2x=x-1
=>-2x-x=-1-5
=>-3x=-6
=>x=2(nhận)
b: ĐKXĐ: \(\begin{cases}x^2-3x+2\ge0\\ x-1\ge0\end{cases}\Rightarrow\begin{cases}\left(x-2\right)\left(x-1\right)>=0\\ x-1\ge0\end{cases}\)
=>(x>=2 hoặc x<=1) hoặc x>=1
=>(x>=2 hoặc x=1)
\(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
=>\(x^2-3x+2=x-1\)
=>(x-1)(x-2)-(x-1)=0
=>(x-1)(x-3)=0
=>\(\left[\begin{array}{l}x-1=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)
c: ĐKXĐ: \(\begin{cases}x^2-3x+2\ge0\\ 5x+2\ge0\end{cases}\)
=>\(\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\ 5x+2\ge0\end{cases}\)
=>(x>=2 hoặc x<=1) và x>=-2/5
=>x>=2 hoặc -2/5<=x<=1
\(\sqrt{x^2-3x+2}-\sqrt{5x+2}=0\)
=>\(\sqrt{x^2-3x+2}=\sqrt{5x+2}\)
=>\(x^2-3x+2=5x+2\)
=>\(x^2-8x=0\)
=>x(x-8)=0
=>x=0(nhận) hoặc x=8(nhận)
d: ĐKXĐ: \(\begin{cases}3x+7\ge0\\ x+1\ge0\end{cases}=>x\ge-1\)
\(\sqrt{3x+7}-\sqrt{x+1}=0\)
=>\(\sqrt{3x+7}=\sqrt{x+1}\)
=>3x+7=x+1
=>2x=-6
=>x=-3(loại)
Bài 8:
a: \(\left|x^2-5x+4\right|=x+4\)
=>\(\begin{cases}x+4\ge0\\ x^2-5x+4=\left(x+4\right)^2\end{cases}\)
=>\(\begin{cases}x\ge-4\\ x^2-5x+4-x^2-8x-16=0\end{cases}\Rightarrow\begin{cases}x\ge-4\\ -13x-12=0\end{cases}\)
=>\(x=-\frac{12}{13}\)
b: \(\left|x^2-7x+12\right|=15-5x\)
=>\(\begin{cases}15-5x\ge0\\ \left(15-5x\right)^2=x^2-7x+12\end{cases}\)
=>\(\begin{cases}5x\le15\\ 25\left(x-3\right)^2=\left(x-3\right)\left(x-4\right)\end{cases}\Rightarrow\begin{cases}x\le3\\ \left(x-3\right)\left(x-4\right)-25\left(x-3\right)^2=0\end{cases}\)
=>\(\begin{cases}x\le3\\ \left(x-3\right)\left(x-4-25x+75\right)=0\end{cases}\Rightarrow\begin{cases}x\le3\\ \left(x-3\right)\left(-24x+71\right)=0\end{cases}\)
=>x=3(nhận) hoặc x=71/24(nhận)
c: \(\left|x^2-6x+5\right|+1=x\)
=>\(\left|x^2-6x+5\right|=x-1\)
=>\(\begin{cases}x-1\ge0\\ x^2-6x+5=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge1\\ x^2-6x+5=x^2-2x+1\end{cases}\)
=>\(\begin{cases}x\ge1\\ -6x+5=-2x+1\end{cases}\Rightarrow\begin{cases}x\ge1\\ -4x=-4\end{cases}\Rightarrow x=1\)
d: \(3x^2+5\left|x-3\right|+7=0\) (1)
TH1: x>=3
(1) sẽ trở thành: \(3x^2+5\left(x-3\right)+7=0\)
=>\(3x^2+5x-15+7=0\)
=>\(3x^2+5x-8=0\)
=>\(3x^2+8x-3x-8=0\)
=>(3x+8)(x-1)=0
=>x=-8/3(loại) hoặc x=1(loại)
TH2: x<3
(1) sẽ trở thành: \(3x^2+5\left(3-x\right)+7=0\)
=>\(3x^2+15-5x+7=0\)
=>\(3x^2-5x+22=0\)
\(\Delta=\left(-5\right)^2-4\cdot3\cdot22=25-12\cdot22<0\)
=>Phương trình vô nghiệm
e: ĐKXĐ: x<>2
\(\frac{x^2-1}{\left|x-2\right|}=x\)
=>\(x^2-1=x\cdot\left|x-2\right|\) (1)
TH1: x>2
(1) sẽ trở thành:
\(x\left(x-2\right)=x^2-1\)
=>\(x^2-2x=x^2-1\)
=>-2x=-1
=>x=1/2(loại)
TH2: x<2
(1) sẽ trở thành: \(x\left(x-2\right)=1-x^2\)
=>\(1-x^2=x^2-2x\)
=>\(x^2-2x+x^2-1=0\)
=>\(2x^2-2x-1=0\)
=>\(x^2-x-\frac12=0\)
=>\(x^2-x+\frac14-\frac34=0\)
=>\(\left(x-\frac12\right)^2=\frac34\)
=>\(\left[\begin{array}{l}x-\frac12=\frac{\sqrt3}{2}\\ x-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt3+1}{2}\left(nhận\right)\\ x=\frac{1-\sqrt3}{2}\left(nhận\right)\end{array}\right.\)
f: \(\frac{\left|x-1\right|}{x^2-x-6}=1\)
=>\(x^2-x-6=\left|x-1\right|\)
=>\(\begin{cases}x^2-x-6\ge0\\ \left(x^2-x-6\right)^2=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}\left(x-3\right)\left(x+2\right)\ge0\\ \left(x^2-x-6-x+1\right)\left(x^2-x-6+x-1\right)=0\end{cases}\)
=>(x>=3 hoặc x<=-2) và \(\left(x^2-2x-5\right)\left(x^2-7\right)=0\)
=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}x^2-2x+1-6=0\\ x^2-7=0\end{array}\right.\)
=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}\left(x-1\right)^2=6\\ x^2=7\end{array}\right.\)
=>(x>=3 hoặc x<=-2) và \(\left[\begin{array}{l}x-1=\sqrt6\\ x-1=-\sqrt6\\ x=\pm\sqrt7\end{array}\right.\)
=>(x>=3 hoặc x<=-2) và x\(\in\left\lbrace\sqrt6+1;-\sqrt6+1;\sqrt7;-\sqrt7\right\rbrace\)
=>\(x\in\left\lbrace\sqrt6+1;-\sqrt7\right\rbrace\)
Bài 6:
a: ĐKXĐ: \(x^2-4<>0\)
=>(x-2)(x+2)<>0
=>x∉{2;-2}
\(\frac{x^2-3x+5}{x^2-4}=-1\)
=>\(x^2-3x+5=-x^2+4\)
=>\(2x^2-3x+1=0\)
=>(x-1)(2x-1)=0
=>\(\left[\begin{array}{l}x-1=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=\frac12\left(nhận\right)\end{array}\right.\)
b: ĐKXĐ: x∉{2;-2/3}
\(\frac{2x+1}{3x+2}=\frac{x+1}{x-2}\)
=>(3x+2)(x+1)=(2x+1)(x-2)
=>\(3x^2+3x+2x+2=2x^2-4x+x-2\)
=>\(3x^2+5x+2-2x^2+3x+2=0\)
=>\(x^2+8x+4=0\)
=>\(x^2+8x+16-12=0\)
=>\(\left(x+4\right)^2=12\)
=>\(\left[\begin{array}{l}x+4=2\sqrt3\\ x+4=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3-4\left(nhận\right)\\ x=-2\sqrt3-4\left(nhận\right)\end{array}\right.\)
c: ĐKXĐ: x∉{2;-3}
\(1+\frac{2}{x-2}=\frac{10}{x+3}-\frac{50}{\left(2-x\right)\left(x+3\right)}\)
=>\(1+\frac{2}{x-2}=\frac{10}{x+3}+\frac{50}{\left(x-2\right)\left(x+3\right)}\)
=>\(\frac{\left(x-2\right)\left(x+3\right)+2\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{10\left(x-2\right)+50}{\left(x-2\right)\left(x+3\right)}\)
=>(x-2)(x+3)+2(x+3)=10(x-2)+50
=>x(x+3)=10(x-2+5)
=>(x+3)(x-10)=0
=>x=-3(loại) hoặc x=10(nhận)
d: ĐKXĐ: x∉{2;-2;-1}
\(\frac{x+1}{x+2}+\frac{x-1}{x-2}=\frac{2x+1}{x+1}\)
=>\(\frac{\left(x+1\right)\left(x-2\right)+\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+1}{x+1}\)
=>\(\frac{x^2-x-2+x^2+x-2}{x^2-4}=\frac{2x+1}{x+1}\)
=>\(\frac{2x^2-4}{x^2-4}=\frac{2x+1}{x+1}\)
=>\(\frac{2x^2-8+4}{x^2-4}=\frac{2x+2-1}{x+1}\)
=>\(\frac{4}{x^2-4}=\frac{-1}{x+1}\)
=>\(-x^2+4=4x+4\)
=>\(-x^2=4x\)
=>\(x^2+4x=0\)
=>x(x+4)=0
=>x=0(nhận) hoặc x=-4(nhận)
