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\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)
\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)
\(=\dfrac{486}{375}=\dfrac{162}{125}\)
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
Theo đề ta có:\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x^2}{64}=\dfrac{y^2}{144}=\dfrac{z^2}{225}\)
Áp dụng t/c của dãy tỉ số = nhau ta có:
\(\dfrac{x^2}{64}=\dfrac{y^2}{144}=\dfrac{z^2}{225}=\dfrac{x^2-y^2}{64-144}=\dfrac{-16}{-80}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{5}\cdot64=\dfrac{64}{5}\\y^2=\dfrac{1}{5}\cdot144=\dfrac{144}{5}\\z^2=\dfrac{1}{5}\cdot225=45\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt{\dfrac{64}{5}};x=-\sqrt{\dfrac{64}{5}}\\y=\sqrt{\dfrac{144}{5}};y=-\sqrt{\dfrac{144}{5}}\\z=\sqrt{45};z=-\sqrt{45}\end{matrix}\right.\)
Vậy............................
\(=\left(\dfrac{3}{2}-3\right).\sqrt{\dfrac{25}{16}}=\left(-\dfrac{3}{2}\right).\dfrac{5}{4}=-\dfrac{15}{8}\)
\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}-\dfrac{3}{2}+1=\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1=-\dfrac{3}{2}\)
= 4 . -1/8 - 2 . -1/4 + 3 . -1/2 + 1
= -1/2 - -1/2 + -3/2 + 1
= -1/2
