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

\(\Leftrightarrow2x^2+2t^2+y^2-t^2=2xy+2xt\)

\(\Leftrightarrow2x^2+2t^2+y^2-t^2-2xy-2xt=0\)

\(\Leftrightarrow x^2+x^2+t^2+y^2-2xy-2xt=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xt+t^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-t\right)^2=0\)

Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-t\right)^2\ge0\end{cases}}\)nên \(\left(x-y\right)^2+\left(x-t\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Leftrightarrow x=y=t\left(đpcm\right)\)

Theo đề, ta có:   \(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}=\dfrac{t}{x}\) \(=\dfrac{x+y+z+t}{y+z+t+x}=1\) .

\(\Rightarrow x=y;y=z;z=t;t=x\)

\(\Rightarrow x=y=z=t\)

\(M=\dfrac{2x-y}{z+t}+\dfrac{2y-z}{t+x}+\dfrac{2z-t}{x+y}+\dfrac{2t-x}{y-z}\)

\(M=\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}\)

\(M=\dfrac{1}{2}.4\)

\(M=2\)

Từ bài ra => 2x^2+2t^2+y^2-t^2 = 2xy + 2xt

<=> 2x^2+t^2+y^2=2xy+2xt

<=>2x^2+y^2+t^2-2xy-2xt=0

<=>(x^2-2xy+y^2)+(x^2-2xt+t^2)=0

<=> (x-y)^2+(x-t)^2 = 0

<=> x-y=0 và x-t=0

<=> x=y=t

=> ĐPCM

A=xy-xz+2z-2y

B=2xy-2xz+2²- yt²

C=xy-2yz+y²

^{_{bạn tự tính kết quả nha}}

a: \(A=\left(y-z\right)\left(x-2\right)\)

\(=\left(2-2\right)\cdot\left(1.007-0.06\right)=0\)

b: \(B=2\cdot18.3\cdot\left(24.6-10.6\right)+\left(2-24.6\right)\left(2+31.7\right)\)

\(=36.6\cdot14-761.62=-249.22\)

c: \(C=\left(x-y\right)\left(y+z\right)-y\left(x-y\right)\)

\(=\left(0.86-0.26\right)\left(0.26+1.5\right)-0.26\left(0.86-0.26\right)\)

\(=0.6\cdot1.5=0.9\)

Phân số cuối cùng chắc em ghi nhầm

\(\dfrac{x}{y+z+t}+\dfrac{y+z+t}{9x}\ge2\sqrt{\dfrac{x\left(y+z+t\right)}{9x\left(y+z+t\right)}}=\dfrac{2}{3}\)

Tương tự:

\(\dfrac{y}{z+t+x}+\dfrac{z+t+x}{9y}\ge\dfrac{2}{3}\)

\(\dfrac{z}{t+x+y}+\dfrac{t+x+y}{9z}\ge\dfrac{2}{3}\)

\(\dfrac{t}{x+y+z}+\dfrac{x+y+z}{9t}\ge\dfrac{2}{3}\)

Đồng thời:

\(\dfrac{8}{9}\left(\dfrac{y+z+t}{x}+\dfrac{z+t+x}{y}+\dfrac{t+x+y}{z}+\dfrac{x+y+z}{t}\right)\)

\(\ge\dfrac{8}{9}\left(\dfrac{3\sqrt[3]{yzt}}{x}+\dfrac{3\sqrt[3]{ztx}}{y}+\dfrac{3\sqrt[3]{txy}}{z}+\dfrac{3\sqrt[3]{xyz}}{t}\right)\)

\(\ge\dfrac{8}{3}.4\sqrt[4]{\dfrac{\sqrt[3]{yzt}.\sqrt[3]{ztx}.\sqrt[3]{txy}.\sqrt[3]{xyz}}{xyzt}}=\dfrac{32}{3}\)

Cộng vế:

\(VT\ge4.\dfrac{2}{3}+\dfrac{32}{3}=\dfrac{40}{3}\)

Dấu "=" xảy ra khi \(x=y=z=t\)

Câu 2: 

\(B=x^2+2x+y^2-2x-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2\cdot7+37=49+37+14=100\)

Câu 3: 

\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2\cdot5+10=25\)

\(\frac{x}{y}=\frac{z}{t}\Rightarrow=\frac{x}{z}=\frac{y}{t}=\frac{2x}{2z}\Rightarrow\frac{2x^2}{2z^2}=\frac{y^2}{t^2}\)

\(\frac{2x^2}{2z^2}=\frac{y^2}{t^2}=\frac{2x^2-y^2}{2z^2-t^2}\)

\(^{\frac{y^2}{t^2}=\frac{y}{t}\cdot\frac{y}{t}=\frac{x}{z}\cdot\frac{y}{t}=\frac{xy}{zt}\left(1\right)}\)

\(\frac{y^2}{t^2}=\frac{2y^2-y^2}{2z^2-t^2}\left(2\right)\)

từ (1) và (2)=>\(\frac{xy}{zt}=\frac{2x^2-y^2}{2z^2-t^2}\left(đpcm\right)\)