Ta có: \(x+y=m+n\Rightarrow n=x+y-m\)

\(\Rightarrow S=x^2+y^2+m^2+\left(x+y-m\right)^2\)

\(=x^2+y^2+m^2+(x^2+y^2+m^2+2xy-2mx-2my)\)

\(=x^2+y^2+m^2+(x^2+y^2+m^2+2xy-2mx-2my)\)

\(=x^2+y^2+m^2+x^2+y^2+m^2+2xy-2mx-2my\)

\(=\left(x^2+2xy+y^2\right)+\left(m^2-2mx+x^2\right)+\left(m^2-2my+y^2\right)\)

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

Vì x, y, m, n \(\in\) Z nên x + y; m - x; m - y là số nguyên

Vậy S luôn bằng tổng các bình phương của 3 số nguyên

bài của   Never_NNL   sai nhé:

  \(x+y=m+n\)   \(\Rightarrow\)\(n=x+y-m\)

Ta có:    \(A=x^2+y^2+m^2+n^2\)

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

\(=2x^2+2y^2+2m^2+2xy-2mx-2my\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-2mx+m^2\right)+\left(y^2-2my+m^2\right)\)

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

Vậy A là tổng của 3 số chính phương

x + y = m + n

m = x + y - n

x^2 + y^2 + ( x + y - n )^2 + n^2 

= x^2 + y^2 + ( x^2 + xy- xn ) + ( xy + y^2 - ny ) - [ ( - xn ) + ( - ny ) + n^2 ] + n^2 

= x^2 + y^2 + x^2 + xy - xn + xy + y^2 - ny + xn + ny - n^2 + n^2 

= 2x^2 + 2y^2 + 2xy 

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

= x^2 + y^2 + ( x + y )^2 ( dpcm )

cảm ơn 2 bạn nha

Ta có: x + y = m + n

\(\Leftrightarrow\) x = m + n - y

Thay vào S ta được:

\(S=x^2+y^2+m^2+n^2\)

\(=\left(m+n-y\right)^2+y^2+m^2+n^2\)

\(=m^2+n^2+y^2+2mn-2ny-2ym+y^2+m^2+n^2\)

\(=\left(m^2+2mn+n^2\right)+\left(n^2-2ny+y^2\right)+\left(y^2-2ym+m^2\right)\)

\(=\left(m+n\right)^2+\left(n-y\right)^2+\left(y-m\right)^2\)

Vậy ...

Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)với a,b>0 

Ta có: \(\frac{4xy}{z+1}=\frac{4xy}{2z+x+y}\le\frac{xy}{x+z}+\frac{xy}{y+z}\)

Tương tự: \(\frac{4yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)

                \(\frac{4zx}{y+1}\le\frac{zx}{y+x}+\frac{zx}{y+z}\)

\(\Rightarrow4\left(\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\right)\le\frac{xy}{x+z}+\frac{xy}{y+z}+\frac{yz}{x+y}+\frac{yz}{x+z}+\frac{zx}{y+x}+\frac{zx}{y+z}=x+y+z=1\)

\(\Rightarrow\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{1}{4}\)

Dấu "=" xảy ra khi: x=y=z>0

Bài 2: 

+) Với y=0 <=> x=0

Ta có: 1-xy= 1² (đúng) 

+) Với \(y\ne0\)

Ta có: \(x^6+xy^5=2x^3y^2\)

\(\Leftrightarrow x^6-2x^3y^2+y^4=y^4-xy^5\)

\(\Leftrightarrow\left(x^3-y^2\right)^2=y^4\left(1-xy\right)\)

\(\Rightarrow1-xy=\left(\frac{x^3-y^2}{y^2}\right)^2\)

Cảm ơn bạn Phạm Quốc Cường.

Đặt \(\left(\dfrac{x}{6};\dfrac{y}{3};\dfrac{z}{2}\right)=\left(a;b;c\right)\Rightarrow2^{6a}+4^{3b}+8^{2c}=4\)

\(\Leftrightarrow64^a+64^b+64^c=4\)

Áp dụng BĐT Cô-si:

\(4=64^a+64^b+64^c\ge3\sqrt[3]{64^{a+b+c}}\Rightarrow64^{a+b+c}\le\dfrac{64}{27}\)

\(\Rightarrow a+b+c\le log_{64}\left(\dfrac{64}{27}\right)\Rightarrow M=log_{64}\left(\dfrac{64}{27}\right)\)

Lại có: \(x;y;z\ge0\Rightarrow a;b;c\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}64^a\ge1\\64^b\ge1\\64^c\ge1\end{matrix}\right.\) \(\Rightarrow\left(64^b-1\right)\left(64^c-1\right)\ge0\)

\(\Rightarrow64^{b+c}+1\ge64^b+64^c\) (1)

Lại có: \(b+c\ge0\Rightarrow64^{b+c}\ge1\Rightarrow\left(64^a-1\right)\left(64^{b+c}-1\right)\ge0\)

\(\Rightarrow64^{a+b+c}+1\ge64^a+64^{b+c}\) (2)

Cộng vế (1);(2) \(\Rightarrow4=64^a+64^b+64^c\le64^{a+b+c}+2\)

\(\Rightarrow64^{a+b+c}\ge2\Rightarrow a+b+c\ge log_{64}2\)

\(\Rightarrow N=log_{64}2\)

\(\Rightarrow T=2log_{64}\left(\dfrac{64}{27}\right)+6log_{64}\left(2\right)\approx1,4\)

\(x+y=m+n\Rightarrow x+y-m-n=0\Rightarrow2x\left(x+y-m-n\right)=0\)

Do đó: \(S=x^2+y^2+m^2+n^2+2x\left(x+y-m-n\right)\)

\(S=x^2+y^2+m^2+n^2+2x^2+2xy-2xm-2xn\)

\(S=\left(x^2+2xy+y^2\right)+\left(x^2-2xm+m^2\right)+\left(x^2-2xn+n^2\right)\)

\(S=\left(x+y\right)^2+\left(x-m\right)^2+\left(x-n\right)^2\)

Vậy \(x^2+y^2+m^2+n^2\) bằng tổng bình phương của ba số nguyên