Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a) Ta có: \(8x+4x^2-12xy\)
\(=4x\left(2+x-3y\right)\)
b) Ta có: \(5x^3-10x^2+5x\)
\(=5x\left(x^2-2x+1\right)\)
\(=5x\left(x-1\right)^2\)
c) Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) Ta có: \(x^2-8x-9\)
\(=x^2-9x+x-9\)
\(=\left(x-9\right)\left(x+1\right)\)
a. `8x+4x^2-12xy=4x(2+x-3y)`
b) `5x^3-10x^2+5x=5x(x^2-2x+1)`
c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`
d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`
\(=x^2\left(x-1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\\ =\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
\(4x^4-8x^3+3x^2-8x+4\)
\(=\left(4x^4-8x^3\right)+\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=4x^3\left(x-2\right)+3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(4x^3+3x-2\right)\)
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
4x2 - 8x + 3
= 4x2 - 6x - 2x + 3
= ( 4x2 - 6x ) - ( 2x - 3 )
= 2x( 2x - 3 ) - ( 2x - 3 )
= ( 2x - 3 )( 2x - 1 )
\(4x^2-8x+3\)
\(=4x^2-2x-6x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-1\right)\left(2x-3\right)\)
\(=4x^2-8x+4-3\)
\(=\left(2x-2-\sqrt{3}\right)\left(2x-2+\sqrt{3}\right)\)
= (2x)3- y3 + ( 4x2y +2xy + y2)
= (2x - y +1)(4x2y + 2xy + y2)
( hiu k? a3 - b3 mà a = 2x ; b =y)
Sửa đề: 4x^2-y^2+8x+4
=4x^2+8x+4-y^2
=(2x+2)^2-y^2
=(2x+y+2)(2x-y+2)