Question

Giải Phương Trình:

\(a.\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)

\(b.\frac{x^2}{x^2+2x+2}+\frac{x^2}{x^2-2x+2}-\frac{4\left(x^2-5\right)}{x^2+4}=\frac{322}{65}\)

 

Answer 1

a/ \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)

<=> \(\frac{\left(x+1\right)^2}{\left(x+1\right)^2+1}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2+2}=\frac{7}{6}\left(1\right)\)

đặt \(\left(x+1\right)^2=a\left(a>0\right)\)

=> \(\left(1\right)\)<=> \(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)

<=> \(\frac{a\left(a+2\right)+\left(a+1\right)^2}{\left(a+1\right)\left(a+2\right)}=\frac{7}{6}\)

<=> \(\frac{2a^2+4a+1}{a^2+3a+2}=\frac{7}{6}\)

<=> \(6\left(2a^2+4a+1\right)=7\left(a^2+3a+2\right)\)

<=> \(5a^2+3a-8=0\)

<=> \(5a^2-5a+8a-8=0\)

<=>  \(\left(5a+8\right)\left(a-1\right)=0\)

<=> \(a=\frac{-8}{5}\left(h\right)a=1\)

mà \(a>0\)

=> \(a=1\)

=> \(\left(x+1\right)^2=1\)

=> \(x+1=1\left(h\right)x+1=-1\)

=> \(x=0\left(h\right)x=-2\)

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Answer 2

Xét x = 0 và x = -2 , thay vào ta được \(VT=VP\)

Xét x > 0 : 

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

\(=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)>\frac{7}{6}=VP\) ( loại ) 

Xét x < -2 : 

\(VT=2-\left(\frac{1}{x\left(x+2\right)+2}+\frac{1}{x\left(x+2\right)+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{7}{6}=VP\) ( loại ) 

Xét -2 < x < 0 : 

\(VT=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{-2}+1\right)=\frac{3}{2}>\frac{7}{6}=VP\) ( loại ) 

Vậy ...