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1 tháng 8 2020

Ta có \(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)

=> \(\frac{9}{25}x\left(1+\frac{3}{5}+1\right)+18\left(\frac{3}{5}+1\right)=x\)

=> \(\frac{117}{125}x+28,8=x\)

=> \(x-\frac{117}{125}x=28,8\)

=> \(\frac{8}{125}x=28,8\)

=> x = 450

Vậy x = 450

1 tháng 8 2020

\(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)

\(x\left(\frac{9}{25}+\frac{9}{25}+\frac{9}{25}\right).\frac{3}{5}+\frac{3}{5}.18+18=x\)

\(x.\frac{3}{5}\left(\frac{27}{25}+18\right)+18=x\)

\(x.\frac{3}{5}\left(\frac{27}{25}+\frac{450}{25}\right)+18=x\)

\(x.\frac{3}{5}.\frac{477}{25}+18=x\)

\(x.\frac{1431}{125}+\frac{2250}{125}=x\)

\(x.\frac{3681}{125}=x\)

vậy chac tui làm sai rồi

2 tháng 4 2017

khó quá 

2 tháng 4 2017

\(a.\frac{-5}{9}+\frac{5}{9}:A=\left(-\frac{5}{9}+\frac{5}{9}\right):A=0:A=0\)

\(b.\frac{7}{25}.\frac{11}{13}-\frac{7}{25}.\frac{2}{13}-\frac{8}{25}=\frac{7}{25}.\left(\frac{11}{13}-\frac{2}{13}\right)-\frac{8}{25}=\frac{7}{25}.\frac{9}{13}-\frac{8}{25}=\frac{63}{325}-\frac{8}{25}=\frac{-41}{325}\)

C.Thua

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6 tháng 6 2016

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

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

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

6 tháng 6 2016

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

24 tháng 3 2019

Bài 1: Tìm x biết:

1) x +\(\frac{7}{12}\)\(\frac{17}{18}\)\(\frac{1}{9}\)                                                                               2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)

 x +\(\frac{7}{12}\)\(\frac{15}{18}\)                                                                                                                \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)\(\frac{7}{69}\)

             x = \(\frac{15}{18}\)\(\frac{7}{12}\)                                                                                                    \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)

             x = \(\frac{1}{4}\)                                                                                                                                   x = \(\frac{3}{10}\)

10 tháng 4 2016

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

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

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

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

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

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_