a) \(x^2-8x+y^2+6y+25=0\)

\(\left(x-8\right)x+y\left(y+6\right)+25=0\)

\(x^2+y^2+6y+25=8x\)

\(\Rightarrow x=4,y=-3\)

b )^​ ​4x²-4x+9y² -12y +5

<=> [( 2x )²​ - 4x + 1 ] [ (3y) ² ​- 12y + 4 )] = 0

<=> ( 2x - 1 )² ​ + ( 3y - 2 )²​ =0   ( Vì (2x -1)² ​>=0 , ( 3y - 2 )² >= 0 )

<=> 2x - 1 = 0 và 3y -2 = 0

<=> x = 1/2     và y = 2/3

toi ko bt

\(x^2+y^2=0\)

Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)

(Dấu "="\(\Leftrightarrow x=y=0\))

Lời giải:
a)

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

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

Với $x=-3; y=\frac{1}{2}$ thì:

$xy=\frac{-3}{2}; x^2=9; y^2=\frac{1}{4}$

Do đó $A=\frac{-35}{46}$

b)
\(B=\frac{(8x^3-y^3)(4x^2-y^2)}{(2x+y)(4x^2-4xy+y^2)}=\frac{(2x-y)(4x^2+2xy+y^2)(2x-y)(2x+y)}{(2x+y)(2x-y)^2}\)

\(=4x^2+2xy+y^2=4.2^2+2.2.\frac{-1}{2}+(\frac{-1}{2})^2=\frac{57}{4}\)

Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)

\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

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

Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(2x-y-z\right)^2=0\\\left(y-3\right)^2=0\\\left(z-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)

Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)

Thay \(\left(1\right)\)vào \(S\),ta được :

\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)

    \(=0-1+1=0\)

Vậy \(S=0\)

Bạn tự tách hđt nhé! Gõ mỏi tay :v~

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

⇔ \(y^2-2yz+z^2+z^2-2xz+x^2+x^2-2xy+y^2=\)\(6(z^2-yz-xz+y^2-xy+x^2)\)

⇔ \(2\left(x^2+y^2+z^2-yz-xz-xy\right)\)=\(6(z^2-yz-xz+y^2-xy+x^2)\)

⇔ \(x^2+y^2+z^2-yz-xz-xy\) = \(3(z^2-yz-xz+y^2-xy+x^2)\)

⇔ \(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

⇔ \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z\)

Do đó \(\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)

⇒ \(x=y=z\)

j lắm thế :)))

Bài 2 : ~ bài 1 ngán quá =)))

a, Có

\(5x^2+10y^2-6xy-4x-2y+3\)

\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)+1\)

\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x;y\)

Do đó không tồn tại x , y tm \(5x^2+10y^2-6xy-4x-2y+3=0\)

b, \(x^2+4y^2+z^2-2x-6x+6y+15=0\)

Câu này đề sai :v bài ngta không cho 2 lần x vậy đâu bạn :)))

Bài 1:

a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)

\(=10-10x=10(1-x)\)

b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)

\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)

\(=-7x^2+7x=7x(1-x)\)

c)

\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)

\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)

\(=\left\{3-x-5[9x-2]\right\}(-2x)\)

\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)

Bài 2:

a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)

\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)

\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)

b)

\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)

\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)

\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)

\(2x^2+3(x^2-1)=5x(x+1)\)