\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.

a) Ta có: \(\frac{x}{9}=\frac{-12}{27}\)

=> \(27.x=-12.9\)

=> \(27x=-108\)

=> \(x=108:27\)

=>\(x=4\)

a)

\(\frac{x-2}{4}=\frac{x+5}{-3}\)

\(\Rightarrow\frac{-3.\left(x-2\right)}{4.\left(-3\right)}=\frac{4.\left(x+5\right)}{4.\left(-3\right)}\)

\(\Rightarrow\frac{-3x+6}{-12}=\frac{4x+20}{-12}\)

\(\Rightarrow-3x+6=4x+20\)

\(\Rightarrow-3x-4x=20-6\)

\(\Rightarrow-7x=14\)

\(\Rightarrow x=-2\)

b)

\(\frac{x}{3}=\frac{27}{x}\)

\(\Rightarrow x.x=3.27\)

\(\Rightarrow x^2=81\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)

T**k mik nhé!

(4-x)^3=27

lên mạng đi 

\(\frac{x+3}{3}=\frac{27}{x+3}\left(ĐKXĐ:x\ne-3\right)\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}\)( Tm ĐKXĐ ) 
Vậy ...

Theo bài ra ta cs (x+3).(x+3)=3.27

=>(x+3)^2=81

=>x+3=9

=>x=6

Vậy...

\(a,\frac{27}{x+1}=\frac{x+1}{3}\)

\(\Leftrightarrow27.3=\left(x+1\right)\left(x+1\right)\)

\(\Leftrightarrow81=\left(x+1\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=9^2\\\left(x+1\right)^2=\left(-9\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}}\)

Vậy \(x\in\left\{8;-10\right\}\)

a)\(\frac{27}{x+1}=\frac{x+1}{3}\)

Ta có : 27 . 3 = (x + 1)(x + 1)

=> 81 = (x + 1)²

=> 9² = (x + 1)²

=> \(\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}\)

\(b,\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)

\(\Rightarrow x=\frac{6}{9}=\frac{2}{3}\)

Vậy \(x=\frac{2}{3}\)

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

1/ \(\Rightarrow\left(\frac{1}{3}\right)^x\left[1+\left(\frac{1}{3}\right)^2\right]=\frac{10}{27}\)

\(\Rightarrow\left(\frac{1}{3}\right)^x.\frac{10}{9}=\frac{10}{27}\)

\(\Rightarrow\left(\frac{1}{3}\right)^x=\frac{1}{3}\Rightarrow\left(\frac{1}{3}\right)^x=\left(\frac{1}{3}\right)^1\Rightarrow x=1\)

2/ Có: 2a + 7b = 28

=> 2a + 2a = 28 (vì 2a = 7b)

=> 4a = 28

=> a = 7

Thay a = 7 vào 2a = 7b ta đc:

2.7 = 7.b

=> b = 2

Vậy a = 7 ; b = 2