ai trả lời được thì tôi k đúng cho :))

thì kệ

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)

Đặt \(a=\frac{x+2}{x-1};b=\frac{x-2}{x+1}\)

=>\(ab=\frac{x+2}{x-1}\cdot\frac{x-2}{x+1}=\frac{x^2-4}{x^2-1}\)

Phương trình trở thành: \(a^2+b^2-6ab=0\)

=>\(a^2-6ab+9b^2=8b^2\)

=>\(\left(a-3b\right)^2=\left(2b\sqrt2\right)^2\)

=>\(\left[\begin{array}{l}a-3b=2b\sqrt2\\ a-3b=-2b\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}a=3b+2b\sqrt2=b\left(3+2\sqrt2\right)\\ a=3b-2b\sqrt2=b\cdot\left(3-2\sqrt2\right)\end{array}\right.\)

TH1: \(a=b\left(3+2\sqrt2\right)\)

=>\(\frac{x+2}{x-1}=\frac{x-2}{x+1}\cdot\left(3+2\sqrt2\right)\)

=>\(\left(x+2\right)\left(x+1\right)=\left(3+2\sqrt2\right)\cdot\left(x-2\right)\left(x-1\right)\)

=>\(\left(3+2\sqrt2\right)\left(x^2-3x+2\right)=x^2+3x+2\)

=>\(x^2\left(3+2\sqrt2\right)-x^2+\left(-9-6\sqrt2-3\right)x+6+4\sqrt2-2=0\)

=>\(x^2\left(2\sqrt2+2\right)+x\left(-12-6\sqrt2\right)+4\sqrt2+4=0\) (1)

\(\Delta=\left(-12-6\sqrt2\right)^2-4\cdot\left(2\sqrt2+2\right)\cdot\left(4\sqrt2+4\right)\)

\(=144+72+144\sqrt2-4\left(16+8\sqrt2+8\sqrt2+8\right)\)

\(=216+144\sqrt2-4\left(24+16\sqrt2\right)=216+144\sqrt2-96-64\sqrt2=120+80\sqrt2=40\left(3+2\sqrt2\right)\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{12+6\sqrt2-\sqrt{40\left(3+2\sqrt2\right)}}{2\left(2\sqrt2+2\right)}=\frac{12+6\sqrt2-2\sqrt{10}\left(\sqrt2+1\right)}{2\left(2\sqrt2+2\right)}=\frac{\left(\sqrt2+1\right)\left(6-2\sqrt{10}\right)}{4\left(\sqrt2+1\right)}=\frac{6-2\sqrt{10}}{4}=\frac{3-\sqrt{10}}{2}\\ x=\frac{12+6\sqrt2+\sqrt{40\left(3+2\sqrt2\right)}}{2\left(2\sqrt2+2\right)}=\frac{12+6\sqrt2+2\sqrt{10}\left(\sqrt2+1\right)}{2\left(2\sqrt2+2\right)}=\frac{\left(\sqrt2+1\right)\left(6+2\sqrt{10}\right)}{4\left(\sqrt2+1\right)}=\frac{6+_{}2\sqrt{10}}{4}=\frac{3+\sqrt{10}}{2}\end{array}\right.\)

TH2: \(a=b\left(3-2\sqrt2\right)\)

=>\(\frac{x+2}{x-1}=\frac{x-2}{x+1}\cdot\left(3-2\sqrt2\right)\)

=>\(\left(x+2\right)\left(x+1\right)=\left(3-2\sqrt2\right)\cdot\left(x-2\right)\left(x-1\right)\)

=>\(\left(3-2\sqrt2\right)\left(x^2-3x+2\right)=x^2+3x+2\)

=>\(x^2\left(3-2\sqrt2\right)-x^2+\left(-9+6\sqrt2-3\right)x+6-4\sqrt2-2=0\)

=>\(x^2\left(-2\sqrt2+2\right)+x\left(-12+6\sqrt2\right)-4\sqrt2+4=0\) (1)

\(\Delta=\left(-12+6\sqrt2\right)^2-4\cdot\left(-2\sqrt2+2\right)\cdot\left(-4\sqrt2+4\right)\)

\(=144+72-144\sqrt2-4\left(16-8\sqrt2-8\sqrt2+8\right)\)

\(=216-144\sqrt2-4\left(24-16\sqrt2\right)=216-144\sqrt2-96+64\sqrt2=120-80\sqrt2=40\left(3-2\sqrt2\right)\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{12-6\sqrt2-\sqrt{40\left(3-2\sqrt2\right)}}{2\left(-2\sqrt2+2\right)}=\frac{12-6\sqrt2-2\sqrt{10}\left(\sqrt2-1\right)}{2\left(-2\sqrt2+2\right)}=\frac{\left(\sqrt2-1\right)\left(6-2\sqrt{10}\right)}{4\left(-\sqrt2+1\right)}=\frac{-6+2\sqrt{10}}{4}=\frac{-3+\sqrt{10}}{2}\\ x=\frac{12-6\sqrt2+\sqrt{40\left(3-2\sqrt2\right)}}{2\left(-2\sqrt2+2\right)}=\frac{12-6\sqrt2+2\sqrt{10}\left(\sqrt2-1\right)}{2\left(-2\sqrt2+2\right)}=\frac{\left(\sqrt2-1\right)\left(6+2\sqrt{10}\right)}{4\left(-\sqrt2+1\right)}=\frac{-6-_{}2\sqrt{10}}{4}=\frac{-3-\sqrt{10}}{2}\end{array}\right.\)