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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a)\(\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)

\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}\)

\(=\frac{-7}{25}.1-\frac{18}{25}\)

\(=-\frac{7}{25}-\frac{18}{25}\)

\(=-1\)

b)\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}\)

\(=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)

\(=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)

\(=\frac{5}{7}.0\)

\(=0\)

\(x\times\frac{6}{25}=\frac{15}{-13}\)

x=\(\frac{15}{-13}\div\frac{6}{25}\)

x=\(-\frac{125}{26}\)

các câu còn lại làm tương tự nha!!!

\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)

\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)

\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)

\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)

\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)

Câu a:

- \(\frac12\)(3 - 2\(x\)) - 7 = 5 - \(\frac13\)(\(x\) - \(\frac45\))

- \(\frac32\) + \(x\) - 7 = 5 - \(\frac13x\) + \(\frac{4}{15}\)

\(x\) + \(\frac13x\) = 5 + \(\frac{4}{15}\) + 7 + \(\frac32\)

(1 + \(\frac13\))\(x\) = \(\frac{150}{30}\) + \(\frac{8}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{158}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{368}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{413}{30}\)

\(x\) = \(\frac{413}{30}\) : \(\frac43\)

\(x\) = \(\frac{413}{40}\)

Vậy \(x=\frac{413}{40}\)

Câu b:

(5 - 3x/2) : - 1 3/8 = - 7 1/3

(5 - 3x/2) : (-11/8) = - 22/3

5 - 3x/2 = - 22/3 x (-11/8)

5 - 3x/2 = 121/12

3x/2 = 5 - 121/12

3x/2 = - 61/12

x = - 61/12 : 3/2

x = -61/18

Vậy x = - 61/18

B)

Câu 1a:

75% - 1 1/2 + 0,5 : 5/12 - (-1/2)^2

= 0,75 - 1,5 + 0,5 x 12/5 - 0,25

= (0,75 - 0,25) - 1,5 + 1,2

= 0,5 - 1,5 + 1,2

= - 1 + 1,2

= 0,2

Bài 1b:

(5/7.0,6 - 5: 3 1/2). (40% - 1,4).(-2)^3

= (3/7 - 5 x 2/7).(0,4 - 1,4).8

= - 1.(-1).8

= 8

a) |-3| - 2.x = -7

<=> 3 - 2.x = -7

<=> -2.x = -7 - 3

<=> -2.x = -10

<=> x = (-10) : (-2)

<=> x = 5

=> x = 5

b) \(x+75\%=\frac{7}{8}\)

\(\Leftrightarrow x+\frac{75}{100}=\frac{7}{8}\)

\(\Leftrightarrow x+\frac{3}{4}=\frac{7}{8}\)

\(\Leftrightarrow x=\frac{7}{8}-\frac{3}{4}\)

\(\Leftrightarrow x=\frac{1}{8}\)