\(\left(\frac{-32}{27}\right)-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)
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LL
1
31 tháng 1 2022
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=-1-\dfrac{5}{27}+\dfrac{24}{27}=-1+\dfrac{19}{27}=-\dfrac{8}{27}\)
=>3x-7/9=-2/3
=>3x=-2/3+7/9=1/9
hay x=1/27
S
3
A
31 tháng 3 2018
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\left(3x-\frac{7}{9}\right)^3=\frac{-32}{27}-\left(-\frac{24}{27}\right)\)
\(\left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=\frac{-2}{3}\)
\(3x=\frac{-2}{3}+\frac{7}{9}\)
\(3x=\frac{1}{9}\)
\(x=\frac{1}{9}:3\)
\(x=\frac{1}{27}\)
TV
1
29 tháng 6 2016
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}--\frac{24}{27}\)
\(\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\left(3x-\frac{7}{9}\right)=\sqrt[3]{-\frac{8}{27}}\)
Rồi làm tiếp đi
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)
b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)
c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)
d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)
e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)
g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)
i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)
k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)
MS
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
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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)\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{24}{27}\)
=\(x=-\frac{35}{27}\)
\(\frac{-32}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)
\(-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}+\frac{32}{27}\)
\(-\left(3x-\frac{7}{9}\right)^3=\frac{8}{27}\)
\(\left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}\)
\(\left(3x-\frac{7}{9}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(3x-\frac{7}{9}=\frac{-2}{3}\)
\(3x=\frac{-2}{3}+\frac{7}{9}\)
\(3x=\frac{1}{9}\)
\(x=\frac{1}{9}:3\)
\(x=\frac{1}{27}\)
\(V\text{ậy}\) \(x=\frac{1}{27}\)
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