a) (9xy - 3x2)(-2y2 - 8xy)
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\(3x^2\left(3x^2-2y^2\right)-\left(3x^2-2y^2\right)\left(3x^2+2y^2\right)\)
\(=9x^4-6x^2y^2-9x^4+4y^4\)
\(=-6x^2y^2+4y^4\)
3: \(x^4-13x^2+36\)
\(=x^4-9x^2-4x^2+36\)
\(=\left(x^2-9\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
4: \(x^4+3x^2-2x+3\)
\(=x^4+x^3+3x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(x^2-x+1\right)\)
5: \(x^4+2x^3+3x^2+2x+1\)
\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)
\(=\left(x^2+x+1\right)^2\)
\(3x^2\cdot\left(3x^2-2y^2\right)-\left(3x^2-2y^2\right)\cdot\left(3x^2+2y^2\right)\)
\(\Leftrightarrow\left(3x^2-2y^2\right)\cdot\left(3x^2-3x^2+2y^2\right)\)
\(\Leftrightarrow2y^2\cdot\left(3x^2-2y^2\right)\)
\(\Leftrightarrow6x^2y^2-4y^4\)
`A= x^2+2xy-3x^2 +2y^2+3x^2-y^2`
`= (x^2-3x^2 +3x^2) +2xy +(2y^2 -y^2)`
`= x^2 +2xy +y^2`
`=(x+y)^2`
A = \(x^2\) + 2\(xy\) - 3\(x^2\) + 2y2 + 3\(x^2\) - y2
A = (\(x^2\)- 3\(x^2\) + 3\(x^2\)) + 2\(xy\) + (2\(y^2\) - y2)
A = \(x^2\) + 2\(xy\) + y2
A = (\(x\) + y)2
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
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
a) Ta có: \(\dfrac{6x^2-8xy}{9xy-12y^2}\)
\(=\dfrac{2x\left(3x-4y\right)}{3y\left(3x-4y\right)}=\dfrac{2x}{3y}\)
b) \(\dfrac{2a^3-18a}{a^4-81}\)
\(=\dfrac{2a\left(a^2-9\right)}{\left(a^2-9\right)\left(a^2+9\right)}=\dfrac{2a}{a^2+9}\)
a) Ta có:
\(A+B\)
\(=3x^2-9xy+y^2-7-y^2-3x+12\)
\(=3x^2-9xy+5-3x\)
b) Ta có:
\(A-B\)
\(=\left(3x^2-9xy+y^2-7\right)-\left(-y^2-3x+12\right)\)
\(=3x^2-9xy+y^2-7+y^2+3x-12\)
\(=3x^2-9xy+2y^2+3x-19\)
a) \(a+b=\left(3x^2-9xy+y^2\right)+\left(-y^2-3x+12\right)\)
\(=3x^2-9xy+y^2-y^2-3x+12\)
\(=3x^2-9xy+\left(y^2-y^2\right)-3x+12\)
\(=3x^2-9xy-3x+12\)
b) \(a-b=\left(3x^2-9xy+y^2\right)-\left(-y^2-3x+12\right)\)
\(=3x^2-9xy+y^2+y^2+3x-12\)
\(=3x^2-9xy+\left(y^2+y^2\right)+3x-12\)
\(=3x^2-9xy+2y^2+3x-12\)
\(\left(9xy-3x^2\right)\left(-2y^2-8xy\right)=-18xy^3-72x^2y^2+6x^2y^2+24x^3y=-18xy^3-66x^2y^2+24x^3y\)
(9xy - 3x2)(-2y2 - 8xy)
= -18xy3 - 72x2y2 + 6x2y2 + 24x3y
= -18xy3 - 66x2y2 + 24x3y