a,\(=-3x\left(x^2+4x+4\right)+\left(x-3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)

     \(=-3x^3-12x^2-12x+x^3-x-3x^2+3-4x^2+12x-9\)

    \(=-2x^3-19x^2-x-3\)

b,\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

     \(=x^3+2x^2-9x-18-x^3+3x+x^2-3-5x^3-40x^2-80x-x^2+10x-25\)

      \(=-5x^3-38x^2-76x-46\)

`1/2 xx 1/3 xx 1/4`

`= (1xx1xx1)/(2xx3xx4)`

`= 1/24`

__

`1/2 xx 1/3 : 1/4`

`= 1/2 xx 1/3 xx 4`

`= (1xx1xx4)/(2xx3)`

`= 4/6`

`=2/3`

__

`1/2 : 1/3 xx1/4`

`= 1/2 xx 3 xx 1/4`

`=(1xx3xx1)/(2xx4)`

`= 3/8`

__

`1/2 : 1/3 : 1/4`

`= 1/2 xx 3xx4`

`= 12/2`

`=6`

`1/2xx1/3xx1/4`

`=1/24`

`1/2xx1/3:1/4`

`=1/6xx4`

`=4/6=2/3`

`1/2:1/3xx1/4`

`=1/2xx3xx1/4`

`=3/2xx1/4`

`=3/8`

`1/2:1/3:1/4`

`=1/2xx3xx4`

`=6`

=6

Tran bang

Bạn đợi tí nha ! Mình đang làm !

Câu d là dấu " ) "  đúng không bạn ?

\(1+x-2+2x=3\)

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

\(-1+3x=3\)

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

\(3x=4\)

\(x=\frac{4}{3}\)

a)1+x-2(1-x)=3

=)1+x-2+2x=3

=)-1+3x=3

=)3x=4

=)x=4/3

\(1,\left(x-3\right).\left(x+4\right)>0\)

<=> x - 3 và x + 4 cùng dấu 

<=> TH1 : 

\(\hept{\begin{cases}x-3>0\\x+4>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>3\\x>-4\end{cases}\Leftrightarrow x>3}}\)

TH2 : 

\(\hept{\begin{cases}x-3< 0\\x+4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x< -4\end{cases}\Leftrightarrow x< -4}}\)

Vậy với x>3 hoặc x<-4 thì ( x-3) . ( x +4 ) >0

\(2,\left(x-5\right).\left(x+7\right)< 0\)

<=> x - 5 và x + 7 khác dấu 

<=> TH1 : 

\(\hept{\begin{cases}x-5>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>5\\x< -7\end{cases}}}\)( vô lí )

TH2 : 

\(\hept{\begin{cases}x-5< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 5\\x>-7\end{cases}\Leftrightarrow-7< x< 5}}\)

Vậy với -7 < x < 5 thì ( x - 5 ) . ( x + 7)<0

\(3,\left(x^2+1\right).\left(x-3\right)>0\)

<=> x^2 + 1 và x -3 cùng dấu 

<=> TH1 : 

\(\hept{\begin{cases}x^2+1>0\\x-3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2>-1\\x>3\end{cases}\Leftrightarrow}x>3}\)

TH2 : 

\(\hept{\begin{cases}x^2+1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2< -1\\x< 3\end{cases}\Leftrightarrow x^2< -1}}\)

Vậy với x> 3 hoặc x^2 < -1 thì ( x^2 + 1 ) .( x - 3 ) >0

       \(27-\left(14-x\right)=-x+3\)

\(\Leftrightarrow\)\(27-14+x=-x+3\)

\(\Leftrightarrow\)\(13+x=-x+3\)

\(\Leftrightarrow\)\(2x=-10\)

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

Vậy...

Câu 1 :

=> 27-14+x = -x+3

=> 13+x = -x+3

=> x = -x+3-13 = -x-10

=> 10 = -x-x = -2x

=> x = 10 : (-2) = -5

Vậy x = -5

Tk mk nha

s cậu k cho coin đi! Cho like làm j?

1. x(x + 7) = 0

=> x = 0

x + 7 = 0 => x =  -7

Vậy x = 0 ; -7

2. (x + 12)(x - 3) = 0

x + 12 = 0 => x = -12

x - 3 = 0 => x = 3

Vậy x = -12 ; 3

3. (-x + 5)(3 - x) = 0

-x + 5 = 0 => -x = -5 => x = 5

3 - x = 0 => x = 3

Vậy z = 5 ; 3

4. x(2 + x)(7 - x) = 0

=> x = 0

2 + x = 0 => x = -2

7 - x = 0 => x = 7

Vậy x = 0 ; -2 ; 7

5. (x - 1)(x + 2)(-x - 3) = 0

x - 1 = 0 => x = 1

x + 2 = 0 => x = -2

-x - 3 = 0 => -x = 3 => x = -3

Vậy x = 1 ; -2 ; -3

\(x+\left(x+7\right)=0\)

+) \(x=0\)

+) \(x+7=0=>x=-7\)

Vậy x=0 hoặc x=-7

\(\left(x+12\right).\left(x-3\right)=0\)

+) \(x+12=0=>x=-12\)

+) \(x-3=0=>x=3\)

Vậy x=-12 hoặc x=3

\(x.\left(2+x\right).\left(7-x\right)=0\)

+) \(x=0\)

+) \(2+x=0=>x=-2\)

+) \(7-x=0=>x=7\)

Vậy x=0 hoặc x=-2 hoặc x=7

\(\left(x-1\right).\left(x+2\right).\left(x+3\right)=0\)

+) \(x-1=0=>x=1\)

+) \(x+2=0=>x=-2\)

+) \(x+3=0=>x=-3\)

Vậy x=1 hoặc x=-2 hoặc x=-3

1) x.(x+7)=0

TH1: x = 0 thì x + 7 = 7

TH2: x + 7 = 0 thì x = -7

2) (x+12).(x-3)=0

TH1: x + 12 = 0 thì x = -12, x - 3 = -15

TH2:x - 3 = 0 thì x = 3, x + 12 = 15

3) (-x+5).(3-x)=0

TH1: ( -x + 5 ) = 0 thì -x = -5 => x = 5, 3 - x = -2

TH2: ( 3 - x ) = 0 thì x = 3 => -x + 5 = 2

4) x.(2+x).(7-x)=0

TH1: x = 0 thì 2 + x = 2, 7 - x = 7

TH2: 2 + x = 0 thì x = -2, 7 - x = 9

TH3: 7 - x = 0 thì x = 7, 2 + x = 9

5) (x-1).(x+2).(-x-3)=0

TH1: x - 1 = 0 thì x = 1, x + 2 = 3, -x - 3 = -4

TH2: x + 2 = 0 thì x = -2, x - 1 = -3, -x - 3 = -1

TH3: -x - 3 = 0 thì x = -3, x - 1 = -4, x + 2 = -1    

e) 1/3+1/3:x=1

1/3:x=1-1/3

1/3:x=2/3

x=1/3:2/3

x=1/2

f) x:3/4+1/4=-2/3

x:3/4=-2/3-1/4

x:3/4=-11/12

x=-11/12.3/4

x=-11/16

g) x:(1/2+1/3)=6/5

x:5/6=6/5

x=6/5.5/6

x=1

a) \(8x^2+27=\left(x-1\right)^3+\left(x+4\right)^3\)

\(\Leftrightarrow8x^3+27=x^3-2x^2+x-x^2+2x-1+x^3+8x^2+16x+4x^2+32x+64\)

\(\Leftrightarrow8x^3+27=2x^3+9x^2+51x+63\)

\(\Leftrightarrow8x^3+27-2x^3-9x^2-51x-63=0\)

\(\Leftrightarrow6x^3-36-9x^2-51x=0\)

\(\Leftrightarrow3\left(2x^3-12-3x^2-17x\right)=0\)

\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)

\(\Leftrightarrow3\left(2x^2+3x-8x-12\right)\left(x+1\right)=0\)

\(\Leftrightarrow3\left[x\left(2x+3\right)-4\left(2x+3\right)\right]\left(x+1\right)=0\)

\(\Leftrightarrow3\left(2x+3\right)\left(x-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x+3=0\\x-4=0\\x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=4\\x=-1\end{cases}}\)

tớ tưởng áp dụng công thức: \(\left(A+B\right)^3=A^3+B^3+3AB\left(A+B\right)\)

và \(\left(A-B\right)^3=A^3-B^3-3AB\left(A-B\right)\)

à hiểu rồi