4. 

Xét biểu thức : \(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}=1^2+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+2\left(\frac{k-\left(k-1\right)-1}{k\left(k-1\right)}\right)=1^2+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+2\left(\frac{1}{k-1}-\frac{1}{k}-\frac{1}{k\left(k-1\right)}\right)=\left(1+\frac{1}{\left(k-1\right)}-\frac{1}{k}\right)^2\)

\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=\left|1+\frac{1}{k-1}-\frac{1}{k}\right|\)

Áp dụng : \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{1}-\frac{1}{2}\)

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

...............................................................

\(\sqrt{1+\frac{1}{2015^2}+\frac{1}{2016^2}}=1+\frac{1}{2015}-\frac{1}{2016}\)

Cộng vế các đẳng thức trên được : \(B=2016-\frac{1}{2016}\)

ý thứ 2 là 8/7 chứ không phải 8/8 các bạn nhé. M đánh nhầm chữ

bài này mk chịu vì mk mới học lớp 5

Ta có:

\(\left(y^2+y+1\right)\left(x^2+x+1\right)\)

\(=x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+x+y+1\)

\(=x^2y^2+x^2+y^2+2xy+2=x^2y^2+3\)

Ta lại có:

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

\(=\left(y-x\right)\left(x+y+1\right)=-2\left(x-y\right)\)

Theo đề bài ta có: (sửa đề luôn)

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

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

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

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

\(=-\frac{2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

Em xin đóng góp cách 2 ạ 

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}\)

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

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

\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{x^3y^3-\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x+y\right)^2}\)

\(=\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{x^3y^3-\left(x^2-xy+y^2\right)+x^2+2xy+y^2}\)

\(=\frac{\left(x-y\right)\left[x^2+y^2-\left(x+y\right)^2\right]}{x^3y^3+3xy}\)

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

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

Do \(\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)

\(gt\Rightarrow y-1=-x\Rightarrow x-1=-y\)

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Leftrightarrow\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}+\frac{2\left(x-y\right)}{x^2y^2-3}=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2-3}=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left(x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1\right)}+\frac{2\left(x-y\right)}{x^2y^2-3}=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+2xy+x^2+y^2+2\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}=\frac{-2\left(x-y\right)}{x^2y^2+3}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(dpcm\right)\)

đặt x² + y² = a; xy = b. khi đó a - b = 1 hay a = b + 1.

ta phải chứng minh x⁴ + y⁴ - x²y² \(\ge\)\(\frac{1}{9}\)hay a² - 3b² \(\ge\)\(\frac{1}{9}\)  (1)

thế a = b + 1 vào (1) ta được 9b² - 9b - 4 \(\le\)0 hay (3b + 1)(3b - 4) \(\le\)0 hay \(\frac{-1}{3}\le b\le\frac{4}{3}\)

ta sẽ chứng minh \(\frac{-1}{3}\le b\le\frac{4}{3}\).

thật vậy

ta có x² + y^2\(\ge\)2xy nên từ giả thiết suy ra xy \(\le\) 1 hay b \(\le\)1 nên b \(\le\)\(\frac{4}{3}\)

mặt khác từ giả thiết ta có (x + y)² - 3xy = 1 nên 3xy + 1  = (x + y)^2\(\ge\)0 hay xy \(\ge\)\(\frac{-1}{3}\)hay b  \(\ge\)\(\frac{-1}{3}\)

từ đó suy ra đpcm.