Phân tích đa thức sau thành nhân tử.
(x + 5) ^ 2 - 4x ^ 2
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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: =x^2+5x-x-5
=(x+5)(x-1)
b: =4x^2-(y-3)^2
=(2x-y+3)(2x+y-3)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\\ =\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
=\(x^4+2x^3+x^2+4x^2+4x-12\)
=\(x^4+2x^3+5x^2+4x-12\)
=\(x^4-x^3+3x^3-3x^2+8x^2+4x-12\)
=\(x^3(x-1)+3x^2(x-1)+4(2x^2+x-3)\)
=\(x^3(x-1)+3x^2(x-1)+4(2x^2-2x+3x-3)\)
=\(x^3(x-1)+3x^2(x-1)+4[2x(x-1)+3(x-1)]\)
=\(x^3(x-1)+3x^2(x-1)+4(x-1)(2x+3)\)
=\((x-1)[x^3+3x^2+4(2x+3)]\)
=\((x-1)(x^3+3x^2+8x+12)\)
\(\left(x^2+x\right)^2+\left(4x^2+4x\right)+4-16\\ =\left(x^2+x+2\right)^2-16\\ =\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)
\(x^2+x-2\) vẫn còn phân tích được nữa bạn nhé.
\(x^2+x-2=\left(x-1\right)\left(x+2\right)\)
\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)
\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)
\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )
\(=\left(\sqrt{3}-2\sqrt{x}\right)\left(3+2\sqrt{x}\right)\)
\(\left(x+5\right)^2-4x^2\\=\left(x+5\right)^2-\left(2x\right)^2\\ =\left[\left(x+5\right)-2x\right]\left[\left(x+5\right)+2x\right]\\ =\left(x+5-2x\right)\left(x+5+2x\right)\\ =\left(-x+5\right)\left(3x+5\right)\)