Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(2x-5\right)\\ =\left(x-3\right)\left(x+3+2x-5\right)\\ =\left(x-3\right)\left(3x-2\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=y\)
\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
a: =(x-y)(5-y)
b: \(=x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
Ta có \(f\left(x\right)=\left(x+1\right)\left(x-3\right)\left(x+5\right)\left(x+9\right)+256\)
\(f\left(x\right)=\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x+9\right)+256\)
\(f\left(x\right)=\left(x^2+6x+5\right)\left(x^2+6x-27\right)+256\)
\(f\left(x\right)=\left[\left(x^2+6x-11\right)^2-256\right]+256\)
\(f\left(x\right)=\left(x^2+6x-11\right)^2\)
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
Ta có x(x+3)(x+2)(x+5)+9= x(x+5).(x+2)(x+3) +9= (x2+5x)(x2+5x+6)+9
Đặt x2+5x+3=a ta được
(a-3).(a+3)+9= a2-9+9=a2
Thay x2+5x+3 vào biểu thức trên ta được
(x2+5x+3)2
Vậy x(x+3)(x+2)(x+5)= (x2+5x+3)2
\(x\left(x+3\right)\left(x+2\right)\left(x+5\right)+9\)
\(=\left(x^2+5x\right)\left(x^2+5x+6\right)+9\)
\(=\left[\left(x^2+5x+3\right)-3\right]\left[\left(x^2+5x+3\right)+3\right]+9\)
\(=\left(x^2+5x+3\right)^2-9+9\)
\(=\left(x^2+5x+3\right)\)