tìm x biết
x2+5x=0
phân tích đa thức sau thành nhân tử
x2-2x-xy+2y
HELP 28 phút nữa thi rồi
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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: \(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+a^2b^2+2xyab+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2y^2+a^2b^2+b^2x^2\)
\(=y^2\left(x^2+a^2\right)+b^2\left(a^2+x^2\right)\)
\(=\left(x^2+a^2\right)\left(y^2+b^2\right)\)
\(=x^2+x-6x+6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x+6\right)\)
\(a,\Leftrightarrow\left(x+3\right)^2-4\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-4x+12\right)=0\\ \Leftrightarrow\left(x+3\right)\left(15-3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(b,=x^2\left(y-1\right)-\left(y-1\right)^2=\left(y-1\right)\left(x^2-y+1\right)\)
c: \(x^2-10x+21=\left(x-3\right)\left(x-7\right)\)
a: \(x^2y+xy^3-xy-y^3\)
\(=xy\left(x-1\right)+y^3\left(x-1\right)\)
\(=y\left(x-1\right)\left(x+y^2\right)\)
\(a) x^2y+xy^3-xy-y^3\\=(x^2y+xy^3)-(xy+y^3)\\=xy(x+y^2)-y(x+y^2)\\=(x+y^2)(xy-y)\\=y(x+y^2)(x-1)\\b)2x^2+5x+8(xem lại đề)\\c)x^2-10x+21\\=x^2-3x-7x+21\\=x(x-3)-7(x-3)\\=(x-3)(x-7)\)
\(a,=xy\left(x+y^2\right)-y\left(x+y^2\right)=y\left(x+y^2\right)\left(x-1\right)\\ c,=x^2-7x-3x+21=\left(x-7\right)\left(x-3\right)\)
\(x^2-4y^2-2x+1=\left(x-1\right)^2-4y^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
\(x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(x^2-2x-xy+2y=\left(x^2-xy\right)-2\left(x-y\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)