mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn 

a) (5 - x) +12 = -25

<-> 5 - x + 12 = -25

<-> 17 - x = - 25

<-> x = 42

b) 12 - 4(x - 2) = -4

<-> 12 - 4x + 8 = -4

<-> 20 - 4x = -4

<-> 4x = 24

<-> x = 6

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

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

ủng hộ mk nha bn

bạn trả lời giúp

a) \(x=\frac{1}{27}\)

b) \(x=\frac{2}{5}\)

Các bạn ơi giúp mk với các bạn ơi mk sắp phải đi học rồi giúp mk với

các bạn có thương mk ko

\(a,\frac{5}{7}+\frac{2}{7}:x=\frac{-11}{14}\)

\(\Rightarrow\frac{2}{7}:x=\frac{-11}{14}-\frac{5}{7}\)

\(\Rightarrow\frac{2}{7}:x=\frac{-11}{14}-\frac{10}{14}\)

\(\Rightarrow\frac{2}{7}:x=\frac{-21}{14}\)

\(\Rightarrow\frac{2}{7}:x=\frac{-3}{2}\)

Tự tìm x được rồi :v 

\(b,\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)

\(\Rightarrow\left[\frac{2}{3}-\frac{1}{2}\right]x=\frac{5}{12}\)

\(\Rightarrow\left[\frac{4}{6}-\frac{3}{6}\right]x=\frac{5}{12}\)

\(\Rightarrow\frac{1}{6}x=\frac{5}{12}\)

\(\Rightarrow x=\frac{5}{12}:\frac{1}{6}=\frac{5}{12}\cdot\frac{6}{1}=\frac{5}{2}\cdot\frac{1}{1}=\frac{5}{2}\)

\(c,\left[x+\frac{1}{2}\right]^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left[x+\frac{1}{2}\right]^2=\frac{26}{25}-\frac{17}{25}\)

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

\(\Rightarrow x+\frac{1}{2}=\pm\frac{3}{5}\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=\frac{3}{5}\\x+\frac{1}{2}=-\frac{3}{5}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{10}\\x=-\frac{11}{10}\end{cases}}\)

d, Tự làm

P/S : Thông cảm cho mk nhé vì máy tính mk ko bấm dấu ngoặc tròn được :v

\(-3\frac{3}{5}.x=\frac{4}{15}\)

\(\frac{-18}{5}.x=\frac{4}{15}\)

\(x=\frac{4}{15}:\frac{-18}{5}\)

\(x=\frac{-2}{27}\)

a)

\(x+\dfrac{1}{3}=\dfrac{7}{26}\\ x=\dfrac{7}{26}-\dfrac{1}{3}=\dfrac{21}{78}-\dfrac{26}{78}=-\dfrac{5}{78}\)

b)

\(\dfrac{x}{150}=\dfrac{5}{6}\cdot\dfrac{-7}{25}\\ \dfrac{x}{150}=\dfrac{-7}{30}\\ x\cdot30=-150\cdot7\\ x=\dfrac{-150\cdot7}{30}=-35\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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