a: |x-1|+|x-2|=|2x-3|

=>|x-1|+|x-2|-|2x-3|=0(1)

TH1: x<1

=>x-1<0; 2x-3<0; x-2<0

(1) sẽ trở thành: 1-x+2-x-(3-2x)=0

=>3-2x-3+2x=0

=>0x=0(luôn đúng)

TH2: 1<=x<3/2

=>x-1>=0; 2x-3<0; x-2<0

(1) sẽ trở thành: x-1+2-x-(3-2x)=0

=>1-3+2x=0

=>2x-2=0

=>x=1(nhận)

TH3: 3/2<=x<2

=>x-1>0; 2x-3>=0; x-2<0

(1) sẽ trở thành: x-1+2-x-(2x-3)=0

=>1-2x+3=0

=>-2x+4=0

=>-2x=-4

=>x=2(loại)

TH4: x>=2

=>x-1>0; 2x-3>0; x-2>=0

(1) sẽ trở thành: x-1+x-2-(2x-3)=0

=>2x-3-2x+3=0

=>0x=0(luôn đúng)

Vậy: x<=1 hoặc x>=2

b: ĐKXĐ: x∉{2;3;5}

\(\frac{1}{x-2}+\frac{2}{x-3}-\frac{3}{x-5}=\frac{1}{x^2-5x+6}\)

=>\(\frac{1}{x-2}+\frac{2}{x-3}-\frac{3}{x-5}=\frac{1}{\left(x-2\right)\left(x-3\right)}=\frac{1}{x-3}-\frac{1}{x-2}\)

=>\(\frac{2}{x-2}+\frac{1}{x-3}-\frac{3}{x-5}=0\)

=>\(\frac{2\left(x-3\right)\left(x-5\right)+\left(x-2\right)\left(x-5\right)-3\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)\left(x-5\right)}=0\)

=>2(x-3)(x-5)+(x-2)(x-5)-3(x-2)(x-3)=0

=>\(2\left(x^2-8x+15\right)+x^2-7x+10-3\left(x^2-5x+6\right)=0\)

=>\(2x^2-16x+30+x^2-7x+10-3x^2+15x-18=0\)

=>-8x+22=0

=>-8x=-22

=>x=11/4(nhận)

minh biet

ta có : 

\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)

\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)

\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)

\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)

1. x(x-3)-(x+2)(x-1)=3 <=> x² - 3x - x² - x + 2 = 3 => 4x = -1 => x = 1/4 

2. 

a) x = 0, x=1 (2 nghiệm, loại)

b) x² + 1 > 0 => x = - 2 (1 nghiệm, chọn b)

c) <=> x(x-3) = 0 => x = 0, x=3 (2 nghiệm, loại)

d) (x-1)² + 2 > 0 => Vô nghiệm (loại)

Bài 3: 

b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)

hay \(x\in\left\{0;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

lx

lỗi r bn