a) \(\left(x-6\right)^{10}=\left(x-6\right)^8\)

\(\Rightarrow\left(x-6\right)^{10}-\left(x-6\right)^8=0\)

\(\Rightarrow\left(x-6\right)^8.\left[\left(x-6\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-6\right)^8=0\\\left(x-6\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-6=0\\\left(x-6\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+6\\x-6=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x-6=1\\x-6=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=7\\x=5\end{matrix}\right.\)

Vậy \(x\in\left\{6;7;5\right\}.\)

Chúc em học tốt!

mày tên gì tau trả lời cho

 Troi a gio hoi ten moi bao bai ha  

Thoi giup to di choc to se bao ten cho

số 14 nha bạn ~~~

\(\frac{1}{6}=\frac{x}{18}\Rightarrow x=3\)

\(\frac{x}{8}=-\frac{1}{4}\Rightarrow x=-2\)

\(\frac{4}{-5}=\frac{x}{10}\Rightarrow x=-8\)

\(\frac{11}{5}=-\frac{22}{x}\Rightarrow x=-10\)

\(\frac{x}{8}=\frac{8}{x}\Rightarrow x^2=64\Rightarrow x=8\)

\(\frac{x}{-11}=-\frac{11}{x}\Rightarrow x^2=121\Rightarrow x=11\)

#H

a) (10-x)²:3+3=-9

    (10-x)²:3=-9-3

    (10-x)²:3=-11

    (10-x)²=-11.3

    (10-x)²=-33

 Vô lí vì -33 ko chuyển đc mũ 2 

Vậy x e 0

b) |x-6|.7-4=31

    |x-6|=(31+4):7

    |x-6|=5

\(\Rightarrow\orbr{\begin{cases}x-6=5\Rightarrow5+6=11\\x-6=-5\Rightarrow x=-5+6=1\end{cases}}\)

Vậy.....................................................

a) \(\frac{\left(10-x\right)^2}{3}+3=-9\)

\(\Leftrightarrow\frac{\left(10-x\right)^2}{3}=-12\)

\(\Leftrightarrow\left(10-x\right)^2=-36\)( vô nghiệm vì (10-x)² \(\ge\)0 với mọi x thuộc Z)

b) |x-6|.7-(-2)²=-8+39

<=> |x-6|.7-4=31

<=> |x-6|.7=35

<=> |x-6|=5

\(\Leftrightarrow\orbr{\begin{cases}x-6=5\\x-6=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=11\\x=1\end{cases}}}\)

a) \(\dfrac{x}{15}\)=\(\dfrac{x-2}{30}\)

=>30x=15 (x-2)

=>30x=15x-30

=>30x-15x=-30

=>15x =-30

=>x =-2

b)\(\dfrac{7}{8}\)<\(\dfrac{x}{80}\)<\(\dfrac{9}{10}\)

=>\(\dfrac{70}{80}\)<\(\dfrac{x}{80}\)<\(\dfrac{72}{80}\)

=> 70<x<72

=>x=71

c)\(\dfrac{1}{6}\)<\(\dfrac{5}{x}\)<\(\dfrac{1}{4}\)

=>\(\dfrac{5}{30}\)<\(\dfrac{5}{x}\)<\(\dfrac{5}{20}\)

=>20<x<30

=>x={21;22;23;24;25;26;27;28;29}

\(2(x-5)-3(x+6)=10+x\)

\(\Rightarrow2x-10-3x+18=10x\)

\(\Rightarrow2x-10-3x-18=10+x\)

\(\Rightarrow-x-28=10+x\)

\(\Rightarrow-x-x=10+28\)

\(\Rightarrow-2x=38\)

\(\Rightarrow x=-19\)

Vậy ..

\(2\left(x-5\right)-\left(3x+6\right)=10+x\)

\(\Rightarrow2x-10-3x-18=10+x\)

\(\Rightarrow-x-28=10+x\)

\(\Rightarrow-x=38+x\)

\(\Rightarrow-2x=38\Leftrightarrow x=-19\)