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

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

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

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

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

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.

x = từ 1 đến 10000....0

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)

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

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

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

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

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 \(x=\frac{2}{3}\)

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

a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)

\(x=\left(\frac{3}{7}\right)^2\)

\(x=\frac{9}{49}\)

Vậy...

b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)

\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)

\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)

\(x=-\frac{1}{3}\)

Vậy...

c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

=>\(x-\frac{1}{2}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{2}\)

\(x=\frac{5}{6}\)

Vậy...

d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)

=>\(x+\frac{1}{4}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{1}{4}\)

\(x=\frac{5}{12}\)

Vậy...

Phù, mãi mới xong, tk cho mk nha bn

a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Leftrightarrow2x=18\)

\(\Leftrightarrow x=9\)

b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)

\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)

\(\Leftrightarrow x=66\)

b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)

\(\left(\frac{1}{3}\right)^{2.x}=\left(\frac{1}{3}\right)^{3.22}\)

\(2x=3.22\)

\(x=33\)

ta có 2/3^x+1 +2/3^x=20/27 suy ra 2/3^x *2/3+2/3^x=20/27

suy ra 2/3^x(2/3+1)=20/27 suy ra 2/3^x*5/3=20/27 suy ra 2/3^x=20/27:5/3=4/9

suy ra2/3^x=2/3^2 suy ra x=2

\(\left(\frac{2}{3}\right)^{x+1}+\left(\frac{2}{3}\right)^x=\frac{20}{27}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^x\left(\frac{2}{3}+1\right)=\frac{20}{27}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^x\frac{5}{3}=\frac{20}{27}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^x=\frac{20}{27}.\frac{3}{5}=\frac{4}{9}\)

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

=> x = 2