a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

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

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo). Viết đề thế này khó đọc lắm.

bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự

Bạn xem lại đề, vế phải đâu rồi bạn?

\(\dfrac{x}{x-1}+\dfrac{x-1}{x}=2\)

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

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

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

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

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

\(\Leftrightarrow\dfrac{1}{x\left(x-1\right)}=0\)

\(\Leftrightarrow1=0x\left(x-1\right)\)

\(\Leftrightarrow1=0\)

vậy phương trình không có nghiệm

ĐKXĐ : \(x\ne\pm1\)

PT : \(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}=\dfrac{x+1-\left(x+2\right)\left(x-1\right)}{x-1}\)

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

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

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy ...

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x-1}{x+1}-\dfrac{x^2+x-2}{x+1}=\dfrac{x+1}{x-1}-x-2\)

\(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)

\(\Leftrightarrow\dfrac{-x^2+1}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)

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

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

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

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

Suy ra: \(-\left(x^2-2x+1\right)-x-1+x^2-x+2x-2=0\)

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

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(nhận)

Vậy: S={2}

ĐKXĐ: \(x\ne\pm1\)

\(\dfrac{x+1}{X-1}-\dfrac{x-1}{X+1}=\dfrac{4}{x^2-1}\)

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

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

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

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\)(loại)

Vậy phương trình vô nghiệm