phân tích đa thức thành nhân tử
a/ 2x-2y-x2+2xy-y2
b/ 9x2+6xy+y2-25
thank trước nha
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\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
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)
a) \(=\left(6x\right)^2-2.6x.1+1=\left(6x-1\right)^2\)
b) \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)
c) \(=\left(3x-y\right)^2-25=\left(3x-y-5\right)\left(3x-y+5\right)\)
d) \(=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
\(A=x^2-y^2+7x+7y\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+7\right)\)
\(B=4x^3-4x^2+x\)
\(=x\left(4x^2-4x+1\right)\)
\(=x\left(2x-1\right)^2\)
\(C=x^2-6xy+9y^2-9\)
\(=\left(x-3y\right)^2-9\)
\(=\left(x-3y-3\right)\left(x-3y+3\right)\)
A=\(x^2+7x+7y-y^2=\left(x^2-y^2\right)+\left(7x+7y\right)=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)=\left(x+y\right)\left(x-y+7\right)\)
B=\(4x^3-4x^2+x=x\left(4x^2-4x+1\right)=x\left(2x-1\right)^2\)
C=\(x^2+9y^2-9-6xy=\left(x^2-6xy+9y^2\right)-9=\left(x-3y\right)^2-3^2=\left(x-3y-3\right)\left(x-3y+3\right)\)
a, 7x - 14
= 7(x-2)
b, 2x - 2y + \(x^2\)- xy
= (2x-2y) + (\(x^2\)-xy)
= 2(x-y) + x(x-y)
= (x-y)(2+x)
c, 6x + 12
= 6(x+2)
\(a,=7\left(x-2\right)\\ b,=2\left(x-y\right)+x\left(x-y\right)=\left(x+2\right)\left(x-y\right)\\ c,=6\left(x+2\right)\\ d,\text{Sai đề}\)
a: \(3xy^2-3x^3-6xy+3x\)
\(=3x\left(y^2-x^2-2y+1\right)\)
\(=3x\left\lbrack\left(y-1\right)^2-x^2\right\rbrack\)
=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: \(-x^3-4xy^2+4x^2y+16x\)
\(=x\left(-x^2-4y^2+4xy+16\right)\)
\(=x\left\lbrack4^2-\left(x^2-4xy+4y^2\right)\right\rbrack\)
\(=x\cdot\left\lbrack4^2-\left(x-2y\right)^2\right\rbrack\)
=x(4-x+2y)(4+x-2y)
d: \(xz-x^2-yz+2xy-y^2\)
\(=z\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=z(x-y)-\(\left(x-y^{}\right)^2\)
=(x-y)(z-x+y)
e: \(4x^2-y^2-6x+3y\)
=(2x-y)(2x+y)-3(2x-y)
=(2x-y)(2x+y-3)
a) \(x^2+2xy+y^2-4=\left(x+y\right)^2-2^2\)
\(=\left(x+y-2\right)\left(x+y+2\right)\)
b) \(x^2-y^2+x+y=\left(x-y\right)\left(x+y\right)+1\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+1\right)\)
c) \(y^2+x^2+2xy-16=x^2+2xy+y^2-16\)
\(=\left(x+y\right)^2-4^2=\left(x+y+4\right)\left(x+y-4\right)\)
a) x2-2x-y2+2y
=(x2-y2)-(2x-2y)
=(x-y)(x+y)-2(x-y)
=(x-y)(x+y-2)
a) \(2x-2y-x^2+2xy-y^2\)
\(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
b) \(9x^2+6xy+y^2-25\)
\(=\left(3x\right)^2+6xy+y^2-25\)
\(=\left(3x+y\right)^2-5^2\)
\(=\left(3x+y+5\right)\left(3x+y-5\right)\)