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a)\(=\left(x^2-\frac{1}{4}\right)\left(4x-1\right)\)
\(=4x^3-x^2-x+\frac{1}{4}\)
b) \(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+2x^2-x-2\)
c) \(=\frac{x^2y^2}{2}\left(4x^2-y^2\right)\)
\(=2x^4y^2-\frac{x^2y^4}{2}\)
a ) \(\left(x+1\right)\left(x-2\right)=x^2-2x+x-2=x^2-x-2\)
b ) \(\left(4x^4y^4-12x^2y^2\right):4x^2y^2=x^2y^2-3\)
c ) \(\frac{3x^2-1}{2x}+\frac{x^2+1}{2x}=\frac{3x^2-1+x^2+1}{2x}=\frac{4x^2}{2x}=2x\)
d ) \(\frac{x^2}{x-1}+\frac{2x}{1-x}+\frac{1}{x-1}=\left(\frac{x^2}{x-1}+\frac{1}{x-1}\right)+\frac{2x}{1-x}\)
\(=\frac{x^2+1}{x-1}+\frac{2x}{1-x}=\frac{x^2+1}{x-1}+\frac{-2x}{x-1}=\frac{x^2+1-2x}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
a) .......=x2-x-2
b) .........=x2y2-3
c) .......=(3x2-1+x2+1)/2x=4x2/2x=2x
d) x2 /(x-1)+(-2x)/(x-1)+1/(x-1)=(x2-2x+1)/(x-1)=(x-1)2/(x-1)=x-1
e)...
x-y=4
=> x2-2xy+y2=16
<=> 106-2xy =16 (vì x2+y2 =106)
=>xy=(106-16)/2=45
ta có x3 -y3 =(x-y)(x2+xy+y2 )
=4(106+45)=604
1a : x = -1
2a : x = 10
còn mấy bài khác mình không biết giải nha
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: \(=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
1.
a, (x-1) . (x+1) . (x+2) = (x2 - 1)(x + 2) = x3 - x + 2x2 - 2
b, \(\frac{1}{2x^2y^2}\). (2x+y) . (2x-y) = \(\frac{1}{2x^2y^2}\).[(2x)2 - y2] =\(\frac{1}{2x^2y^2}\)(4x2 - y2) = \(\frac{4x^2-y^2}{2x^2y^2}\)
c, (x-1/2). (x+1/2) . (4x-1) = \(\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=4x^3-x-x^2+\frac{1}{4}\)
sửa lại câu a cho hokage naruto
a) ( x - 1 ) . ( x + 1 ) . ( x + 2 )
= x^2 + x - x - 1 . ( x + 2 )
= ( x^2 - 1 ) . ( x + 2 )
= x^3 + 2x^2 - x - 2
= x^2 + 2x^2 - 2
( chỉ góp ý câu a vậy thôi ) các câu khác ko có ý kiến