phân tích thành nhân tử:x2-(a+b)xy+aby2
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a: \(x^2-4xy+4y^2-2x+4y-35\)
\(=\left(x^2-4xy+4y^2\right)-\left(2x-4y\right)-35\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)-35\)
\(=\left(x-2y\right)^2-7\left(x-2y\right)+5\left(x-2y\right)-35\)
\(=\left(x-2y\right)\left(x-2y-7\right)+5\left(x-2y-7\right)\)
\(=\left(x-2y-7\right)\left(x-2y+5\right)\)
c: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+a^2b^2+2xyab+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)
\(=y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)
\(=\left(x^2+a^2\right)\left(y^2+b^2\right)\)
Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
\(=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)
Ta có: \(x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x-5\right)\left(x+3\right)\)
x2 + 4z2 - 4t2 - 4xt
= x2 - 4xt - 4t2 + 4z2
= 4t2 - 4xt + x2 + 4z2
= (2t - x)2 + 4z2
= \(-\left[\left(2t-x\right)^2-4z^2\right]\)
= \(-\left(2t-x-4z\right)\left(2t-x+4z\right)\)
Lm sao bn ra \(\left(2t-x\right)^2+4z^2=-\left[\left(2t-x\right)^2-4z^2\right]\) hay z?
\(=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\)
\(x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+4=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\)(vô lí)
Vậy pt vô nghiệm
`x^2-2x-4y^2+4y`
`=(x^2-4y^2)-2x+4y`
`=(x-2y)(x+2y)-2(x-2y)`
`=(x-2y)(x+2y-2)`
x2-10x+16=x2-8x-2x+16=(x2-8x)-(2x-16)=x(x-8)-2(x-8)=(x-8)(x-2)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(x^2-\left(a+b\right)xy+aby^2=x^2-axy-bxy+aby^2\)
\(=\left(x^2-axy\right)-\left(bxy-aby^2\right)=x\left(x-ay\right)-by\left(x-ay\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)