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\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(\frac{2}{5}-3x\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow\frac{2}{5}-3x=\frac{3}{5}\)
\(\Rightarrow3x=\frac{-1}{5}\)
\(\Rightarrow x=\frac{-1}{15}\)
(2/5 - 3 x ) ^2 = ( 3 / 5 )^ 2
-> 2/5 - 3x = 3/5
3x= 2/5 - 3/5
3x=-1/5
x= -1/5 : 3
x=-1/15
\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\)
\(\left(\frac{2}{5}-3x\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\frac{2}{5}-3x=\frac{3}{5}\)
\(3x=\frac{2}{5}-\frac{3}{5}\)
\(3x=-\frac{1}{5}\)
\(x=-\frac{1}{5}:3\)
\(x=-\frac{1}{15}\)
a)
\(\begin{array}{l}A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right)\\A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\frac{{ - 23}}{5}\\A = \frac{{5.\left( { - 3} \right).11.\left( { - 23} \right)}}{{11.23.5.5}}\\A = \frac{3}{5}\end{array}\)
b)
\(\begin{array}{l}B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\\B = \frac{{13}}{{25}}.\left( {\frac{{ - 7}}{9} - \frac{2}{9}} \right)\\B = \frac{{13}}{{25}}.(-1)\\B = \frac{{-13}}{{25}}.\end{array}\)
Ta có: \(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\frac{\left(1,2\cdot0,5\right)}{\frac{4}{5}}\)
\(=\frac{\frac{4}{5}:\left(\frac{4}{5}\cdot\frac{5}{4}\right)}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right)\cdot\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right)\cdot\frac{36}{17}}+\frac{6}{5}\cdot\frac{1}{2}\cdot\frac{5}{4}\)
\(=\frac{\frac{4}{5}}{\frac{3}{5}}+\frac{\frac{7}{4}}{\frac{119}{36}\cdot\frac{36}{17}}+\frac{3}{4}\)
\(=\frac{4}{5}\cdot\frac{5}{3}+\frac{7}{4}\cdot\frac{1}{7}+\frac{3}{4}=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}=\frac{7}{3}\)
\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}=\left(\frac{\pm3}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{5}-3x=\frac{3}{5}\\\frac{2}{5}-3x=\frac{-3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{15}\\x=\frac{1}{3}\end{cases}}\)
Vậy,............
(2/5–3x)2=9/25
(2/5–3x)2=(3/5)2
==> 2/5–3x=3/5 hoặc 2/5–3x=—3/5
3x=2/5–3/5 hoặc 3x=2/5–(—3)/5
3x=—1/5 hoặc 3x=1
x=—1/5:3 hoặc x=1:3
x=—1/5.1/3 hoặc x=1/3
x=—1/15 hoặc x=1/3