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)

Bạn kiểm tra lại đề nhé

lx

lỗi r bn

a: =>(x^2+x)^2-2(x^2+x)+(x^2+x)-2=0

=>(x^2+x-2)(x^2+x+1)=0

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

=>x=-2 hoặc x=1

b: ĐKXĐ: x<>4; x<>1

PT =>\(\dfrac{x+3+3x-12}{x-4}=\dfrac{6}{1-x}\)

=>(4x-9)(1-x)=6(x-4)

=>4x-4x^2-9+9x=6x-24

=>-4x^2+13x-9-6x+24=0

=>-4x^2+7x+15=0

=>x=3(nhận) hoặc x=-5/4(nhận)

Câu 2:

TH1: \(x^2-3\ge0\)

=>\(x^2\ge3\)

=>\(\left[\begin{array}{l}x\ge\sqrt3\\ x\le-\sqrt3\end{array}\right.\) (1)

BPT sẽ trở thành: \(x^2-3+2x\ge0\)

=>\(x^2+2x-3\ge0\)

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

=>x>=1 hoặc x<=-3

Kết hợp (1), ta được: x>=\(\sqrt3\) hoặc x<=-3

TH2: \(x^2-3<0\)

=>\(x^2<3\)

=>\(-\sqrt3 (2)

BPT sẽ trở thành: \(3-x^2+2x\ge0\)

=>\(x^2-2x-3\le0\)

=>(x-3)(x+1)<=0

=>-1<=x<=3

Kết hợp (2), ta được: -1<=x<\(\sqrt3\)

\(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\\ \Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\\ \Leftrightarrow6x^2+2-6x^2-6x-6=0\\ \Leftrightarrow-6x-4=0\\ \Leftrightarrow x=-\dfrac{2}{3}\)

\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)

\(\Leftrightarrow x^2+x-x^2+3x=4x\)

\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)

\(\Leftrightarrow0x=0\)

Phương trình có vô số nghiệm , trừ x = -1,x = 3

Vậy ...

\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)

\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)

\(\Leftrightarrow12x+1< 36x+4-24x-3\)

\(\Leftrightarrow12x+1< 12x+1\)

\(\Leftrightarrow12x-12x< 1-1\)

\(\Leftrightarrow0x< 0\)

Vậy S = {x | x \(\in R\)}