Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2) Bạn làm phép chia đa thức cho đa thức, kẻ hẳn dấu chia ra như tiểu học ấy. Được kết quả là \(\left(4y^2+1\right)\) dư (-2y+6) nhé.
3) a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
b) \(\left(x^2+1\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow x^2+1=0\) hoặc x-3=0 hoặc x+2=0
Trường hợp 1 loại vì \(x^2\) không âm, hai trường hợp còn lại tìm được x=3 và x = -2.
4) a)\(x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
b) \(5x^2-10xy-20z^2+5y^2\)
= \(5\left(x^2-2xy-4z^2+y^2\right)\)
= \(5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
= 5 ( x-y-2z ) ( x-y+2z )
5) \(x^3=x\Leftrightarrow x=\pm1\)
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
a: \(=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x\cdot5}{4x\left(2x+1\right)}=\dfrac{10}{2x+1}\)
b: \(=\left(\dfrac{1}{x^2+1}+\dfrac{x-2}{x+1}\right):\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{x+1+x^3+x-2x^2-2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x^3-2x^2+2x-1}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
a) \(\dfrac{2x-2y}{x^2-2xy+y^2}=\dfrac{2\left(x-y\right)}{\left(x-y\right)^2}=\dfrac{2}{x-y}\)
b) \(\dfrac{2-2a}{a^3-1}=-\dfrac{2-2a}{1-a^3}=-\dfrac{2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{-2}{\left(1+a+a^2\right)}\)
c) \(\dfrac{x^2-6x+9}{x^2-8x+15}\)
\(=\dfrac{x^2-2.x.3+3^2}{x^2-3x-5x+15}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x^2-3x\right)-\left(5x-15\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)-5\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\)
\(=\dfrac{x-3}{x-5}\)
d) \(\dfrac{x^4-2x^3}{2x^4-x^3}\)
\(=\dfrac{x^3\left(x-2\right)}{x^3\left(2x-1\right)}\)
\(=\dfrac{x-2}{2x-1}\)
a , x(3-x)-y(y-2x)+y(y-2)-x(2y-x)
=3x-x2-y2+2xy+y2-2y-2xy+x2
=3x-(-x2+x2)+(2xy-2xy)-2y
=3x-2y
b,(x2-7)(x+2)-(2x-1)(x-14)+x(x2-2x-22)+35
=x3+2x2-7x-14-2x2+29x-14+x3-2x2-22x+35
=(x3+x3)+(2x2-2x2-2x2)-(-7x+29x-22x)-14-14+35
=2x3-2x2+7
c,(2x+y)(4x2-2xy+y2)-8x3-y3
=2xy2-4x2y+8x3+y3-2xy2+4x2y-8x3-y3
=(2xy2-2xy2)-(-4x2y+4x2y)+(8x3-8x3)+(y3-y3)
=0