pặc pặc....pặc pặc...........pặc pặc......

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Sửa đề :

\(5x^2+5y^2-8xy-2x-2y+2=0\)

\(\Leftrightarrow4x^2+x^2+4y^2+y^2-8xy-2x-2y+1+1=0\)

\(\Leftrightarrow\left(4x^2-8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}2x-2y=0\\x-1=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\x=1\\y=1\end{cases}\Leftrightarrow x=y=1}}\)

Vậy....

dg bận nên mk ghi kq thôi từ kq bn suy ra hạng tử r` pt nhé

Min=-2 khi (x,y)=(1,-1)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

2:

a: \(3xy^2-3x^3-6xy+3x\)

\(=3x\cdot\left(y^2-2y+1-x^2\right)\)

\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)

=3x(y-1-x)(y-1+x)

b: \(3x^2+11x+6\)

\(=3x^2+9x+2x+6\)

=3x(x+3)+2(x+3)

=(x+3)(3x+2)

c: \(-x^3-4xy^2+4x^2y+16x\)

\(=x\left(16+4xy-4y^2-x^2\right)\)

\(=x\cdot\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)

=x(4-x+2y)(4+x-2y)

d: \(xz-x^2-yz+2xy-y^2\)

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

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

=(x-y)(z-x+y)

e: \(4x^2-y^2-6x+3y\)

=(2x-y)(2x+y)-3(2x-y)

=(2x-y)(2x+y-3)

f: \(x^4-x^3-10x^2+2x+4\)

\(=x^4+2x^3-2x^2-3x^3-6x^2+6x-2x^2-4x+4\)

\(=\left(x^2+2x-2\right)\left(x^2-3x-2\right)\)

g: \(\left(x^3-x^2+x\right)\left(121-25y^2-10y\right)-\left(x^3-x^2+x\right)-\left(121-25y^2-10y\right)+1\)

\(=\left(x^3-x^2+x\right)\left(121-25y^2-10y-1\right)-\left(121-25y^2-10y-1\right)\)

\(=\left(x^3-x^2+x-1\right)\left\lbrack121-\left(25y^2+10y+1\right)\right\rbrack\)

\(=\left(x-1\right)\left(x^2+1\right)\left\lbrack121-\left(5y+1\right)^2\right\rbrack\)

=(x-1)(x^2+1)(11-5y-1)(11+5y+1)

=(x-1)(x^2+1)(10-5y)(12+5y)

=5(2-y)(x-1)(x^2+1)(5y+12)

Tham khảo:

CHÚC EM HỌC TỐT NHÁ 

\(5x^2+4x+2x^3+x^4-12=0\)

\(\Leftrightarrow x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^3+2x^2+x^2+2x+6x+12\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[x^2+2\times\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+6\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vì \(\left(x^2+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\)

\(\Rightarrow\left(x^2+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) vô nghiệm

Vậy phương trình có tập nghiệm là\(S=\left\{1;-2\right\}\)

Mình có giải ở câu hỏi trước rồi nhé.