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22 tháng 7 2017

\(8x^3-1=\left(2x\right)^3-1^3=\left(2x-1\right)\left[\left(2x\right)^2+1.2x+1^2\right]\)

22 tháng 7 2017

8x3-1

=24-1

=23

Có VNEN ko bạn

3 tháng 9 2017

bạn viết tắc kiểu gì mà mình ko hiểu

10 tháng 10 2017

Ok , no broblem

9 tháng 10 2017

Bài 1 . ( 20x4y - 25x2y2 - 3x2 y) : 5x2y

= 5x2y.( 4x2 - 5y - \(\dfrac{3}{5}\)) : 5x2y

= 4x2 - 5y - \(\dfrac{3}{5}\)

Bài 2 . a) ( -2x5 + 3x2 - 4x3) : 2x2

= 2x2.( -x3 + \(\dfrac{3}{2}\) - 2x ) : 2x2

= - x3 - 2x + \(\dfrac{3}{2}\)

b) ( x3 - 2x2y + 3xy2) : ( \(\dfrac{1}{2}x\))

= \(\dfrac{1}{2}x\).( 2x2 - 4xy + 6y2) : ( \(\dfrac{1}{2}x\))

= 2x2 - 4xy + 6y2

c) ( 3x2y2 + 6x2y3 - 12xy ) : 3xy

= 3xy.( xy + 2xy2 - 4 ) : 3xy

= xy + 2xy2 - 4

4 tháng 9 2017

a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)

b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)

c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)

d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)

\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)

e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)

a: \(A=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)\)

\(=2^{X2}+3x-10x-15-2x^2+6x\)

=-x-15

b: \(B=\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)\)

\(=48x^2-12x-20x+5+3x-48x^2-7+112x\)

\(=83x-2\)

5 tháng 7 2017

\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)

\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)

\(A=-12x\)

\(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)

\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)

\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)

\(B=-3x-12\)

Câu C tương tự.

Chúc bạn học tốt!!!

5 tháng 7 2017

A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)

A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)

A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)

A = \(\left(-2\right)\left(2x^2+4x+2x+4\right)+4x^2+8\)

A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)

A = \(-4x^2-12x-8+4x^2+8=-12x\)

b) B = \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)

B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)

B = \(-6x^2-3x-6\)

a: \(9x^2-6x+3\)

\(=\left(9x^2-6x+1\right)+2\)

\(=\left(3x-1\right)^2+2\ge2\)

b: \(6x-x^2+1\)

\(=-\left(x^2-6x-1\right)\)

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left(x-3\right)^2+10\le10\)

a: \(x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

=>x=1 hoặc x=3

b: \(x^2+x-12=0\)

=>(x+4)(x-3)=0

=>x=3 hoặc x=-4

c: \(3x^2+2x-5=0\)

\(\Leftrightarrow3x^2+5x-3x-5=0\)

=>(3x+5)(x-1)=0

=>x=1 hoặc x=-5/3

d: \(x^4-2x^2-3=0\)

\(\Leftrightarrow x^4-3x^2+x^2-3=0\)

\(\Leftrightarrow x^2-3=0\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

14 tháng 6 2017

a)\(5x\left(x-3\right)-4x\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow5x^2-15x-4x^2-4x=x^2-4\)

\(\Leftrightarrow x^2-19x-x^2+4=0\)

\(\Leftrightarrow4-19x=0\Leftrightarrow19x=4\Leftrightarrow x=\dfrac{4}{19}\)

b)\(3x\left(x-5\right)+\left(2x+1\right)\left(x-3\right)=5x\left(x-1\right)\)

\(\Leftrightarrow3x^2-15x+2x^2-5x-3=5x^2-5x\)

\(\Leftrightarrow5x^2-20x-3-5x^2+5x=0\)

\(\Leftrightarrow-15x-3=0\)\(\Leftrightarrow-15x=3\Leftrightarrow x=-\dfrac{1}{5}\)

14 tháng 6 2017

a, \(5x\left(x-3\right)-4x\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow5x^2-15x-4x^2-4x=x^2-4\)

\(\Leftrightarrow x^2-19x=x^2-4\)

\(\Leftrightarrow19x=4\)

\(\Leftrightarrow x=\dfrac{4}{19}\)

Vậy...

b, \(3x\left(x-5\right)+\left(2x+1\right)\left(x-3\right)=5x\left(x-1\right)\)

\(\Leftrightarrow3x^2-15x+2x^2-6x+x-3=5x^2-5x\)

\(\Leftrightarrow5x^2-20x-3=5x^2-5x\)

\(\Leftrightarrow-20x-3=-5x\)

\(\Leftrightarrow-15x=3\)

\(\Leftrightarrow x=\dfrac{-1}{5}\)

Vậy...