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5 tháng 5 2021

éo tính được phép tính lên mặt thì tính cái j

a: =(x^3-8y^3)-(x-2y)

=(x-2y)(x^2+2xy+4y^2)-(x-2y)

=(x-2y)(x^2+2xy+4y^2-1)

b: =2(x^3-4x^2-12x+27)

=2[(x+3)(x^2-3x+9)-4x(x+3)]

=2(x+3)(x^2-7x+9)

30 tháng 8 2023

cảm ơn nhiều

 

20 tháng 2 2016

a)3x^3-8x^2-2x+4

=3x^3-2x^2-6x^2+4x-6x+4

=x^2(3x-2)-2x(3x-2)-2(3x-2)

=(x^2-2x-2)(3x-2).đến đây cậu tự làm nha

b)x^3-4x^2+7x-6

=x^3-2x^2-2x^2+4x+3x-6

=x^2(x-2)-2x(x-2)+3(x-2)

=(x-2)(x^2-2x+3)

.đến đây cậu tự làm nha

c)2x^3-9x+2

=2x^3-4x^2+4x^2-8x-x+2

=2x^2(x-2)+4x(x-2)-(x-2)

=(x-2)(2x^2+4x-1)

.đến đây cậu tự làm nha

20 tháng 1 2017

a, 3x2 - 8x2 - 2x+3=0

2x(3-8) - 2x+3=0

2x5 - 2x+3=0

2x5 - 2x=0-3=

2x5 - 2x=-3

2x(5-x)=-3

5-x=-3/2

5-x=1,5

x=5-1,5

x=3,5

22 tháng 1 2017

3,5 nha bn

chúc bn học tốt

happy new year

27 tháng 10 2021

b: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-5\end{matrix}\right.\)

a. (3x - 1)2 - (x + 3)2 = 0

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

\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

 

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

1 tháng 10 2021

`a)x^3-8x^2+16x`

`=x(x^2-8x+16)`

`=x(x-4)^2`

`b)x^2+4y^2+2x-4y-4xy-24`

`=(x-2y)^2+2(x-2y)-24`

`=(x-2y)^2-4(x-2y)+6(x-2y)-24`

`=(x-2y-4)(x-2y+6)`

`c)x^4+x^3-x^2-2x-2`

`=x^4-2x^2+x^3-2x+x^2-2`

`=x^2(x^2-2)+x(x^2-2)+x^2-2`

`=(x^2-2)(x^2+x+1)`

1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)

2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)

3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)

4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)

6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)

7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)

8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)

9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)

10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)

\(=\dfrac{7x+y}{x}\)

 

a: \(\Leftrightarrow8x^2+16x+14x+7=0\)

=>(2x+1)(8x+7)=0

=>x=-1/2 hoặc x=-7/8

b: \(=x^3-x-6x-6\)

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)

5 tháng 1 2022

\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)

a: Ta có: \(x^4-2x^3+2x-1\)

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

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

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

b: Ta có: \(-a^4+a^3+2a^3+2a^2\)

\(=-a^2\left(a^2-a-2a-2\right)\)

c: Ta có: \(x^4+x^3+2x^2+x+1\)

\(=x^4+x^3+x^2+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^2+1\right)\)

26 tháng 10 2019