(x-2)^2+(x+1)^2+2(x-2)(-x-1)

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

=(x-2-x-1)^2

=(-3)^2=9

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

\(\Leftrightarrow3+x^2-x-2=x^2-4+3x+3\)

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

=>-4x=-2

hay x=1/2(nhận)

\(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\left(ĐKXĐ:x\ne1;x\ne-1\right)\) 

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

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

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

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

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

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy phương trình có nghiệm là x=0

mong có thể giúp bạn

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

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

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

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

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

1) ĐKXĐ: \(x\notin\left\{1;-2\right\}\)

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

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

Suy ra: \(2x^2+4x-x+1=2\left(x^2+x-2\right)\)

\(\Leftrightarrow2x^2+3x+1=2x^2+2x-4\)

\(\Leftrightarrow2x^2+3x+1-2x^2-2x+4=0\)

\(\Leftrightarrow x+5=0\)

hay x=-5(thỏa ĐK)

Vậy: S={-5}

2) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

Ta có: \(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)

\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(x+x^2+5x-x-5=x-5\)

\(\Leftrightarrow x^2+5x-5-x+5=0\)

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

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

Vậy: S={0;-4}

a/ ĐKXĐ : \(x\ne1;-2\)

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

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

\(\Leftrightarrow2x^2+3x-x+1=2x^2+4x-2x-4\)

\(\Leftrightarrow2x+1=2x-4\)

\(\Leftrightarrow1=-4\left(loại\right)\)

Vậy...

b/ĐKXĐ :  \(x\ne\pm5\)

\(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)

\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow x+x^2+5x-x-5=x-5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy...

\(A=x^2+4x+4-x^2+6x-9+3x^2-3x=3x^2+7x-5\\ B=4x^2-4x+1+3x^2-12-7x^2=-4x-11\)