Bạn kiểm tra lại đề nhé

lx

lỗi r bn

lx

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)

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

ĐKXĐ : x ≠ 1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 6

pt <=> \(\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{3x^2-9x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)

<=> \(\frac{6x^2-22x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)

=> \(\left(x-6\right)\left(6x^2-22x+18\right)=6\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

(bạn tự khai triển rút gọn nhé)

<=> \(6x^3-58x^2+150x-108=6x^3-36x^2+66x-36\)

<=>\(6x^3-58x^2+150x-108-6x^3+36x^2-66x+36=0\)

<=> \(-22x^2+84x-72=0\)

<=> \(11x^2-42x+36=0\)

(pt này lên lớp 9 mới học nên mình dừng tại đây)

1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)

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

\(ĐK:x\ne3;-2\)

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

\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)

\(\Leftrightarrow x^2+4x+4+x^2-3x-x^2-6=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)

\(\Leftrightarrow\)    \(\left(x+1\right)^2+x-1=x^2+4\) hoặc   \(\left(x+1\right)^2+1-x=x^2+4\)

Xét \(\left(x+1\right)^2+x-1=x^2+4\)

\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)

\(\Leftrightarrow3x-4=0\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

Xét \(\left(x+1\right)^2+1-x=x^2+4\)

\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)

2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)

\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)

\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)

\(\Leftrightarrow-x< 4\)

\(\Leftrightarrow x>4\)

Vậy \(S=\left\{x|x>4\right\}\)

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