Lời giải:

$x^2-2x+y^2+4y+5+(2z-3)^2=0$

$\Leftrightarrow (x^2-2x+1)+(y^2+4y+4)+(2z-3)^2=0$

$\Leftrightarrow (x-1)^2+(y+2)^2+(2z-3)^2=0$

Vì $(x-1)^2\geq 0; (y+2)^2\geq 0; (2z-3)^2\geq 0$ với mọi $x,y,z$

Do đó để tổng của chúng bằng $0$ thì $(x-1)^2=(y+2)^2=(2z-3)^2=0$

$\Leftrightarrow x=1; y=-2; z=\frac{3}{2}$

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

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

x²-2xy+2y²+4y+4+(2z-3)²=0

(x²-2xy+y²)+(y²+4y+4)+(2z-3)²=0

(x-y)²+(y+2)²+(2z-3)²=0

=>x-y=y+2=2z-3=0

=>z=3/2

y=-2

x=-2

\(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)-2z\left(x-y\right)+z+\left(x^2-8x+16\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y-z=0\\x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=1\\x=4\\y=3\end{cases}}\)

Vậy \(x=4\), \(y=3\), \(z=1\)

\(x^2-2x+y^2+4y+5+\left(2z-3\right)^2=0\)

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

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

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\\\left(2z-3\right)^2\ge0\end{cases}}\) nên \(\left(x-1\right)^2+\left(y+2\right)^2+\left(2z-3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\\\left(2z-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\\z=\frac{3}{2}\end{cases}}}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{y-2x+4z}{2x}=\frac{z-2y+4x}{2y}=\frac{x-2z+4y}{2z}=\)\(=\frac{\left(y-2x+4z\right)+\left(z-2y+4x\right)+\left(x-2z+4y\right)}{2x+2y+2z}=\frac{3\left(x+y+z\right)}{2\left(x+y+z\right)}=\frac{3}{2}\)

\(\Rightarrow\left\{\begin{matrix}2\left(y-2x+4z\right)=6x\\2\left(z-2y+4x\right)=6y\\2\left(x-2z+4y\right)=6z\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}y-2x+4z=3x\\z-2y+4x=3y\\x-2z+4y=3z\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}y+4z=5x\\z+4x=5y\\x+4y=5z\end{matrix}\right.\)

\(P=\left(2+\frac{x}{2y}\right)\left(2+\frac{y}{2z}\right)\left(2+\frac{z}{2x}\right)\)

\(P=\frac{4y+x}{2y}.\frac{4z+y}{2z}.\frac{4x+z}{2x}=\frac{5z}{2y}.\frac{5x}{2z}.\frac{5y}{2x}=\frac{125}{8}\)

Vì bài dài nên mình sẽ tách ra nhé.

1a. Ta có:

$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$

$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$

$=-3(-z)(-x)(-y)=3xyz$

$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$

------------------------

$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$

$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$

$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$

$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$

$=-z^5+5xyz^3-5x^2y^2z$

$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$

$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$

Từ $(1);(2)$ ta có đpcm.

1b.

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$

$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$

Do đó:

$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$

$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$

$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$

$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$

$=7xyz(x^2y^2-2xyz^2+z^4)$

$=7xyz(xy-z^2)$

$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$

$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$

$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)

1c. Sử dụng kq phần a,b:

\(10(x^7+y^7+z^7)=70xyz(xy+yz+xz)^2\)

\(=-35xyz(xy+yz+xz).-2(xy+yz+xz)=-35xyz(x+y+z)(x^2+y^2+z^2)\)

\(=\frac{7}{6}.-30xyz(xy+yz+xz)(x^2+y^2+z^2)=\frac{7}{6}.6(x^5+y^5+z^5).(x^2+y^2+z^2)\)

\(=7(x^5+y^5+z^5)(x^2+y^2+z^5)\)

(đpcm)

1d. Áp dụng kq phần a
$6(x^5+y^5+z^5)=-30xyz(xy+y+xz)=15xyz.-2(xy+yz+xz)=15xyz(x^2+y^2+z^2)$

$\Rightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ (đpcm)