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

 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,(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

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

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

a) \(x=\frac{1}{27}\)

b) \(x=\frac{2}{5}\)

sửa nha chỗ 24/7 thay = \(-\frac{24}{7}\)

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

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

\(\left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

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

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

\(3x=\frac{1}{9}\)

\(x=\frac{1}{9}:3\)

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

(3x-7/9)³ = -32/27-(-24/27)

(3x-7/9)³ = -8/27

(3x)³-(7/9)³ = -8/27

(3x³)-343/729 = -8/27

(3x)³ = -8/27+ -343/729

(3x)³ = -559/729

3³ *x³ = -559/729

9*x³ = -559/729

x³ = -559/729: 9

x³ = -559/6561

x = -0,4400282812

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

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

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

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

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

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.

\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)

\(-\frac{5}{6}\times x=\frac{5}{2}\)

\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)

\(x=-3\)

b.

\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)

\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)

\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)

\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)

\(3x-\frac{37}{10}=-\frac{19}{2}\)

\(3x=-\frac{19}{2}+\frac{37}{10}\)

\(3x=\frac{-95+37}{10}\)

\(3x=-\frac{58}{10}\)

\(3x=-\frac{29}{5}\)

\(x=-\frac{29}{5}\div3\)

\(x=-\frac{29}{5}\times\frac{1}{3}\)

\(x=-\frac{29}{15}\)

c.

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21-16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}\times\frac{4}{3}\)

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

d.

\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)

\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)

\(-\frac{2}{3}\times x=\frac{3-2}{10}\)

\(-\frac{2}{3}\times x=\frac{1}{10}\)

\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)

\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)

\(x=-\frac{3}{20}\)

e.

\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)

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

\(\left|x\right|=\frac{20+9}{12}\)

\(\left|x\right|=\frac{29}{12}\)

\(x=\pm\frac{29}{12}\)

Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)

f.

\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)

\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)

\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)

\(2x-\frac{1}{3}=\pm\frac{1}{6}\)

                \(2x=\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{1+2}{6}\)

                \(2x=\frac{3}{6}\)

                \(2x=\frac{1}{2}\)

                  \(x=\frac{1}{2}\div2\)

                  \(x=\frac{1}{2}\times\frac{1}{2}\)

                  \(x=\frac{1}{4}\)

                \(2x=-\frac{1}{6}+\frac{1}{3}\)

                \(2x=\frac{-1+2}{6}\)

                \(2x=\frac{1}{6}\)

                 \(x=\frac{1}{6}\div2\)

                 \(x=\frac{1}{6}\times\frac{1}{2}\)

                 \(x=\frac{1}{12}\)

Vậy x = 1/4 hoặc x = 1/12.

Chúc bạn học tốt

Sorry nha, mik chép lộn đềLàm lại câu a nha

a.

\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{12}\)

\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)

\(-\frac{5}{6}\times x=\frac{5}{12}\)

\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)

\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)

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

Chúc bạn học tốt

\(a,\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)

\(\frac{4}{6}x-\frac{9}{6}x=\frac{5}{12}\)

\(\frac{-5}{6}x=\frac{5}{12}\)

\(x=\frac{5}{12}:\frac{-5}{6}\)

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

\(b,\frac{2}{5}+\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}\)

\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)

\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{4}{10}\)

\(3x-3,7=\frac{-57}{10}:\frac{3}{5}\)

\(3x=\frac{-19}{2}+\frac{37}{10}\)

\(3x=\frac{-95}{10\text{​​}\text{​​}\text{​​}}+\frac{37}{10}\)

\(x=\frac{-29}{5}:3\)

\(x=\frac{-29}{15}\)

ủng hộ mk nha bn

bạn trả lời giúp

tk cho mk nhé mình đang bị âm nek