\(ĐKXĐ:\)   \(\forall x\in Z\)

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

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

\(\Leftrightarrow\)\(\frac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}=\frac{322}{65}\)

\(\Leftrightarrow\)\(\frac{2x^4+10}{x^4+4}=\frac{322}{65}\)

\(\Rightarrow\)\(65\left(2x^4+10\right)=322\left(x^4+4\right)\)

\(\Leftrightarrow\)\(130x^4+650=322x^4+1288\)

\(\Leftrightarrow\)\(192x^4=-638\)  (vô lý)

Vậy pt vô nghiệm

P/S:mk lm bừa thôi, đúng thì you tham khảo, sai thì báo mk biết nha

\(\Leftrightarrow\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)

\(\Leftrightarrow130x^4+1300=322x^4+1288\)

\(\Leftrightarrow-192x^4=-12\)

\(\Leftrightarrow x^4=\dfrac{1}{16}\)

=>x=1/2 hoặc x=-1/2

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 \(a>0\)

=> \(a=1\)

=> \(\left(x+1\right)^2=1\)

=> \(x+1=1\left(h\right)x+1=-1\)

=> \(x=0\left(h\right)x=-2\)

vậy  ......

chúc bn học tốt

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 ... 

bt2.

A=[2(4x^2+4x+5)-2]/(4x^2+4x+5)

=2-2/[(4x+1)^2+4]

A>=2-2/4=3/2

khi x=-1/4

\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)

\(< =>2x-7=0< =>x=\dfrac{7}{2}\)

b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)

\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)

\(< =>-25x-11=0\)

\(< =>x=-0,44\)

a: =>4x^2-24x+36-4x^2+4x-1<10

=>-20x<10-35=-25

=>x>=5/4

b: =>x(x^2-25)-x^3-8<=3

=>x^3-25x-x^3-8<=3

=>-25x<=11

=>x>=-11/25