ĐK:\(x\ne-1;-3;-5;-7;-9\)

\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-...-\frac{1}{x+9}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)\(\Leftrightarrow\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow2\left(x+1\right)\left(x+9\right)=40\)\(\Leftrightarrow x^2+10x-11=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+11=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}\) (thoả)

Vậy....

sai đề bài bn ak

Đầu bài không liên qan bạn ơi

Với: \(x;y\le1\)

Chứng minh rằng: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{2}{1+xy}\)

\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{1+xy}\right)\le0\)

\(\Leftrightarrow\frac{1+xy-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}+\frac{1+xy-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}\le0\)

\(\Leftrightarrow\frac{-x\left(x-y\right)-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)+\left(1+y^2\right)}+\frac{y\left(x-y\right)\left(1+x^2\right)}{\left(1+y^2\right)\left(1+xy\right)\left(1+x^2\right)}\le0\)

\(\Leftrightarrow\left(x-y\right)\left(-x+y-xy^2+x^2y\right)\le0\)

\(\Leftrightarrow\left(x-y\right)^2\left(xy-1\right)\ge\)(Luôn đúng \(\forall x,y\le1\))

\(\RightarrowĐPCM\)

P/s: Sai đâu thì sửa nhé!

Với: x;y ≤ 1 Chứng minh rằng:

1 + x 2 1 + 1 + y 2 1 ≤ 1 + xy 2 ⇔ 1 + x 2 1 − 1 + xy 1 + 1 + y 2 1 − 1 + xy 1 ≤ 0

⇔ 1 + x 2 1 + xy 1 + xy − 1 + x 2 + 1 + y 2 1 + xy 1 + xy − 1 + y 2 ≤ 0

⇔ 1 + x 2 1 + xy + 1 + y 2 −x x − y − 1 + x 2 + 1 + y 2 1 + xy 1 + x 2 y x − y 1 + x 2 ≤ 0

⇔ x − y −x + y − xy 2 + x 2 y ≤ 0