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

Đặt \(x^2+2x+2=t\)Do \(x^2+2x+2>0\forall x\Rightarrow t>0\)

\(\Rightarrow\frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}\)

\(\Leftrightarrow\frac{\left(t+1\right)\left(t+2\right)}{t\left(t+1\right)\left(t+2\right)}+\frac{2t\left(t+2\right)}{t\left(t+1\right)\left(t+2\right)}=\frac{6t\left(t+1\right)}{t\left(t+1\right)\left(t+2\right)}\)

\(\Leftrightarrow t^2+3t+2+2t^2+4t=6t^2+6t\)

\(\Leftrightarrow3t^2-t-2=0\)

\(\Leftrightarrow3t^2-3t+2t-2=0\)

\(\Leftrightarrow\left(3t+2\right)\left(t-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3t+2=0\\t-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-\frac{2}{3}\left(ktm\right)\\t=1\left(tm\right)\end{cases}}}\)

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

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

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

Vậy ...

Đặt (x-1)²+1=a

\(\rightarrow\)\(\frac{1}{a}+\frac{2}{a}+1=\frac{6}{a}+2\)

\(\rightarrow\) (a+1)(a+2)+2a(a+2)=6a(a+1) (phàn mẫu bỏ đi vì là phép cộng và cả 3 cùng mẫu)

\(\rightarrow\) a²+2a+a+2+2a²+4a=6a²+6a

\(\rightarrow\)-3a²+a+2=0

\(\rightarrow\)-3a²-3a+2a+2=0

\(\Rightarrow\)-3a(a+1)+2(a+1)=0

\(\rightarrow\)(-3a+2)(a+1)=0

\(\rightarrow\)a=\(\frac{2}{3}\), a= -1

Với a=\(\frac{2}{3}\) \(\Rightarrow\) (x-1)²+1=\(\frac{2}{3}\)\(\Rightarrow\)x=rỗng

Với a= -1 =>(x-1)²+1=-1 =>x=rỗng

vậy ko có giá trị nào của x 

@Châu's ngốc

\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\1-\frac{1}{x+3}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm3\\x\ne-2\end{cases}}}\)

a ) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\frac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+3-1}{x+3}\)

\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)

\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(=\frac{3.\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)

\(=\frac{3}{x-3}\) 

b ) \(B=-\frac{3}{5}\Leftrightarrow\frac{3}{x-3}=-\frac{3}{5}\)

\(\Leftrightarrow x-3=-5\Leftrightarrow x=-2\) ( do \(x\ne\pm3;x\ne-2\) ) 

c ) \(B< 0\Leftrightarrow\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow\) \(\hept{\begin{cases}x< 3\\x\ne-2\\x\ne-3\end{cases}}\)

không bạn nha

x²+2>0 r

x(x²+2)=0

=> x=0

hai pt trên không tương đương