\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)

\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)

`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`

`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`

`<=>(x+20)(1/12+1/11+1/10)=0`

Vì `1/12+1/11+1/10 ≠ 0`

`=>x+20=0`

`=>x=0-20`

`=>x=-20`

a) 2y - 12y = 0

\(\Rightarrow\) y ( 2-12) = 0

\(\Rightarrow\) y . (-10) =0 

\(\Rightarrow\) y = 0 : (-10) = 0

b) (y-7)(y-8) = 0

\(\Rightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}\Rightarrow}\orbr{\begin{cases}y=7\\y=8\end{cases}}}\)

c) x + x.2+x.3+x.4+...+x.10 = 165

\(\Rightarrow\) x ( 1+2+3+.....+8+9+10) = 165

\(\Rightarrow\)x . \(\frac{\left(1+10\right).10}{2}\)=165

\(\Rightarrow\) x . 55 = 165

\(\Rightarrow x=\frac{165}{55}=3\)

Can you k for me ,Lê Thị Kim Chi!

a) \(2y-12y=0\)

\(\Leftrightarrow-10y=0\)

\(\Leftrightarrow y=0:\left(-10\right)\)

\(\Leftrightarrow y=0\)

b) \(\left(y-7\right)\left(y-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=7\\y=8\end{cases}}\)

c) \(x+x.2+x.3+......+x.10=165\)

\(\Leftrightarrow x.\left(1+2+3+.....+10\right)=165\)

\(\Leftrightarrow x.55=165\)

\(\Leftrightarrow x=165:55\)

\(\Leftrightarrow x=3\)

x⁸.x²-x⁸=0

x⁸(x²-1)=0

x⁸=0 hoac x²-1=0

x⁸=0

x=0

x²-1=0

x²=1

x=1;x=-1

a) => x - 8 = 0 hoặc x³ + 8 = 0

+) x - 8 = 0 => x = 8

+) x³ + 8 = 0 => x³ = - 8 = (-2)³ => x = -2

Vậy x = 8; -2

b) => 4x - 3 - x - 5 = 30 - 3x

=> 3x - 8 = 30 - 3x

=> 3x + 3x = 30 + 8

=> 6x = 38 => x = 38/6 = 19/3

Vậy x = 19/3

a) => x - 8 = 0 hoặc x³ + 8 = 0

+) x - 8 = 0 => x = 8

+) x³ + 8 = 0 => x³ = - 8 = (-2)³ => x = -2

Vậy x = 8; -2

b) => 4x - 3 - x - 5 = 30 - 3x

=> 3x - 8 = 30 - 3x

=> 3x + 3x = 30 + 8

=> 6x = 38 => x = 38/6 = 19/3

Vậy x = 19/3

Ta có :\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

=> \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

=> \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

=> \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

=> x + 20 = 0

=> x = -20

Vậy x = -20

\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

\(\Leftrightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

\(\Rightarrow x+20=0\Rightarrow x=-20\)

cái bên vế phải chả hiểu là j ?? Tớ tạm đặt là    \(\alpha\)    nhé      (   \(\alpha\)là 1 hằng số    )

pt <=>   \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=\alpha\)

<=>   \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=\alpha\)

<=>   \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=\alpha\)

<=>   \(\frac{\left(x+20\right).181}{660}=\alpha\)

<=>   \(x=\frac{660\alpha}{181}-20\)

Bạn tự thay    \(\alpha\)là 1 số nào tùy ý cũng sẽ tìm ngay ra x thôi 

a)(x-8)(x³+8)=0

<=>x-8=0 hoặc x³+8=0

<=>x=8 hoặc x³=-8

<=>x=8 hoặc x=-2

b)(4x-3)-(x+5)=3(10-x)

<=>4x-3-x-5=30-3x

<=>(4x-x)+(-3-5)=30-3x

<=>3x-8=30-3x

<=>6x=38

<=>x=\(\frac{38}{6}=\frac{19}{3}\)

a)

x³ + 8 = 0                                     x - 8 = 0

x³        = -8                                     x      = 8

x        =-2

a) (x-8)(x^3+8)=0

=>x-8=0 hoặc x^3+8=0

x-8=0               x^3+8=0             

x   =0+8           x^3=0-8

x   =0               x^3=-8

                        x^3=-2^3

                        x=-2

=>x=8 hoặc x= -2

a)(x-8)*(x^3+8)=0

nên x-8=0 hoặc x^3+8=0

       x=0+8          x^3=0-8

       x=8       hoặc x^3=-8

                              x=-2

vậy x=8 hoặc x=-2

b)(4x-3)-(x+5)=3(10-x)

     4x-3-x-5=30-3x(quy tắc bỏ ngoặc)

    3x-8=30-3x

          3x+3x=30+8(chuyển vế)

           6x=38

             x=38:6

x=19/3