K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
23 tháng 11 2017
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
23 tháng 11 2017
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
23 tháng 11 2017
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
LT
1
25 tháng 6 2016
(x-1)(x-2)(x+4)(x+5)-72=[(x-1)(x+4)][x-2)(x+5)]-72=(x^2+3x-4)(x^2+3x-10)-72
Đặt x^2+3x-4=t nên x^2+3x-10=t-6. Thay vào (*) ta được :
(x-1)(x-2)(x+4)(x+5)=t.(t-6)-72=t^2-6t-72=t^2-6t+9-81=(t-3)^2-9^2=(t-3-9)(t-3+9)=(t-12)(t+6)=(x^2+3x-16)(x^2+3x+2)
VP
3
NC
4 tháng 10 2020
Ta có: \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left[\left(x-2\right)\left(x-5\right)\right]\cdot\left[\left(x-3\right)\left(x-4\right)\right]+1\)
\(=\left(x^2-7x+10\right)\cdot\left(x^2-7x+12\right)+1\)
\(=\left[\left(x^2-7x+11\right)-1\right]\cdot\left[\left(x^2-7x+11\right)+1\right]\)
\(=\left(x^2-7x+11\right)^2-1+1\)
\(=\left(x^2-7x+11\right)^2\)
4 tháng 10 2020
\(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)
\(=\left(x-2\right)\left(x-5\right)\left(x-4\right)\left(x-3\right)+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)
Đặt t = \(x^2-7x\)
\(t\left(t+2\right)+1\)
\(=t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=\left(x^2-7x+1\right)^2\)
30 tháng 8 2016
x^5 - x^4 - 1
= x^5 - x^3 - x² - x^4 + x² + x + x^3 - x - 1
= x²( x^3 - x - 1 ) - x( x^3 - x - 1 ) + ( x^3 - x - 1 )
= ( x² - x + 1)( x^3 - x - 1 )
x5 + x4 + 1
= x5 - x3 - x2 - x4 + x2 + x + x3 - x - 1
= x2 ( x3 - x - 1) - x ( x3 - x - 1) + 1 ( x3 - x - 1)
= ( x3 - x - 1) ( x2 - x + 1 )