Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(2x-5\right)\\ =\left(x-3\right)\left(x+3+2x-5\right)\\ =\left(x-3\right)\left(3x-2\right)\)
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)
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)
\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
1/\(\left(x^2-25\right)^2-\left(x-5\right)^2\)
<=>\(\left[\left(x-5\right)\left(x+5\right)\right]^2-\left(x-5\right)^2\)
<=>\(\left(x-5\right)^2\left[\left(x+5\right)^2-1\right]\)
2/\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
<=>\(\left[\left(2x-5\right)\left(2x+5\right)\right]^2-9\left(2x-5\right)^2\)
<=>\(\left(2x-5\right)\left[\left(2x+5\right)^2-9\right]\)
#hoctot<3#
\(9\left(x-3y\right)^2-25\left(2x+y\right)^2\)
\(=\left[3\left(x-3y\right)\right]^2-\left[5\left(2x+y\right)\right]^2\)
\(=\left(3x-9y\right)^2-\left(10x+5y\right)^2\)
\(=\left[3x-9y+10x+5y\right]\left[3x-9y-\left(10x+5y\right)\right]\)
\(=\left(13x-4y\right)\left(-7x-14y\right)\)
\(=-7\left(x+2y\right)\left(13x-4y\right)\)
9(x - 3y)² - 25(2x + y)²
= 3².(x - 3y)² - 5².(2x + y)²
= (3x - 9y)² - (10x + 5y)²
= (3x - 9y - 10x - 5y)(3x - 9y + 10x + 5y)
= (-7x - 14y)(13x - 4y)
= -7(x + 2y)(13x - 4y)
\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
phân tích đa thức thành nhân tử x^2-(2x+3)(x-5)+3
= -(x-9)(x+2)
nha bạn chúc bạn học tốt ạ
Ta có : (2x + 5)2 - (x - 9)2
= (2x + 5 - x + 9)(2x + 5 + x - 9)
= (x + 14)(3x - 4)