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

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

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

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

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

=\(x=-\frac{35}{27}\)

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

 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) \(x=\frac{1}{27}\)

b) \(x=\frac{2}{5}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=-1-\dfrac{5}{27}+\dfrac{24}{27}=-1+\dfrac{19}{27}=-\dfrac{8}{27}\)

=>3x-7/9=-2/3

=>3x=-2/3+7/9=1/9

hay x=1/27

a,(5/8/17+-4/17):x+33/182=4/11

=5/4/17:x+33/182=4/11

5/4/17:x=4/11-33/182

5/4/17:x=365/2002

x=5/4/17:365/2002

x=28/4438/6205

b,-1/5/27-(3x-7/9)^3=-24/27

(3x-7/9)^3=-1/5/27--24/27

(3x-7/9)^3=-8/27

(3x-7/9)^3=(-2/3)^3

3x-7/9=-2/3

3x=-2/3+7/9

3x=1/9

x=1/9:3

x=1/27

1/27 k nha!

\(\frac{1}{27}nha\)

AI TK MÌNH, MÌNH TK LẠI

ủng hộ mk nha bn

bạn trả lời giúp

tk cho mk nhé mình đang bị âm nek

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

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

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

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

\(\left(3x-\frac{7}{9}\right)=\sqrt[3]{-\frac{8}{27}}\)

Rồi làm tiếp đi

a) \(\frac{x}{5}-\frac{x}{6}=\frac{3}{10}\\ \frac{6x}{30}-\frac{5x}{30}=\frac{3\cdot3}{10\cdot3}\\ \frac{x}{30}=\frac{9}{30}\\ \Rightarrow x=9\) Vậy x = 9

b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\\ \frac{-32}{27}+\frac{24}{27}=\left(3x-\frac{7}{9}\right)^3\\ \left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}\\ \left(3x-\frac{7}{9}\right)^3=\left(\frac{-2}{3}\right)^3\\ \Rightarrow3x-\frac{7}{9}=\frac{-2}{3}\\ 3x=\frac{-2}{3}+\frac{7}{9}\\ 3x=\frac{-6}{9}+\frac{7}{9}\\ 3x=\frac{1}{9}\\ x=\frac{1}{9}:3\\ x=\frac{1}{9\cdot3}\\ x=\frac{1}{27}\)Vậy \(x=\frac{1}{27}\)