Ta có

(2x−8)2≥0∀x|x−2y|≥0∀x,y}(2x−8)2≥0∀x|x−2y|≥0∀x,y}

Mà 

Dấu "" xảy ra khi

        {(2x−8)2=0|x−2y|=0{(2x−8)2=0|x−2y|=0

Vậy 

mik ko bít

I don't now

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a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

cac ban oi ai xong truoc mk k cho nhe

a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)

\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)

\(\Rightarrow2y-1=3+y\)

\(2y-y=3+1\)

\(y=4\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{2}{3}\)

c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

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

\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)

\(8x^3-1=16x^4-1\)

\(16x^4-8x^3=0\)

\(8x^3\left(2x-1\right)=0\)

Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)

Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)

Vậy x=0 và x=1/2

2x²y - x² -2y - 2 = 0

=>2x²y-x²-2y+1 = 3

=>(2x²y-x²)-(2y-1)=3

=>x²(2y-1)-(2y-1)=3

=>(x²-1)(2y-1)=3

=>x²-1 và 2y-1 thuộc Ư(3)={3;1;-1;-3}

Xét x²-1=3 =>x²=4 =>x=±2 =>2y-1=1 =>y=1

Xét x²-1=1 =>x²=2 (Loại vì x,y nguyên)

Xét x²-1=-1 =>x²=0 =>x=0 =>2y-1=-3 =>y=-1

Xét x²-1=-3 =>x²=-2 (Loại vì bình phương 1 số luôn \(\ge\)0>-2)

Vậy với x=±2 thì y=1 với x=0 thì y=-1

⇒11x+2⇒11x+2 nguyên ⇒x+2=Ư(11)⇒x+2=Ư(11)

Mà x nguyên dương 

Vậy 

\(2x^2y-x^2-2y-2=0\Leftrightarrow x^2\left(2y-1\right)-\left(2y-1\right)-3=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2y-1\right)=3=\left(-1\right)\left(-3\right)=\left(-3\right)\left(-1\right)=1.3=3.1\)

Tới đây giải nghiệm nguyên như bình thường