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\(4\cdot\left(\frac{1}{4}\right)^2+25\cdot\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(=4\cdot\frac{1}{16}+25\cdot\left[\left(\frac{3}{4}\div\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(=\frac{1}{4}+25\cdot\left(\frac{3}{5}\right)^3\div\left(\frac{3}{2}\right)^3\)
\(=\frac{1}{4}+25\cdot\left(\frac{2}{5}\right)^3\)
\(=\frac{1}{4}+25\cdot\frac{8}{125}\)
\(=\frac{1}{4}\cdot\frac{8}{5}\)
\(=\frac{2}{5}\)
\(4.\left(\frac{1}{4}\right)^2+25.\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3\)
\(=4.\frac{1}{16}+25\left[\left(\frac{3}{4}:\frac{5}{4}\right)^3:\right]:\left(\frac{3}{2}\right)^3\)
\(=\frac{1}{4}+25.\left(\frac{3}{5}\right)^3:\left(\frac{3}{2}\right)^3\)
\(=\frac{1}{4}+25.\left(\frac{2}{5}\right)^3\)
\(=\frac{1}{4}+25.\frac{8}{125}\)
\(=\frac{1}{4}+\frac{8}{5}\)
\(=\frac{2}{5}\)
a) \(\frac{81}{16}\)
b) \(\frac{-31}{8}\)
c) \(\frac{2417}{2401}\)
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a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
Ta có:
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Leftrightarrow x=305\)
\(3xyz^2+\left(-\frac{4}{8}\right)xyz^5\cdot\frac{1}{3}xyz\)
\(=3xyz^2-\frac{1}{2}xyz\cdot\frac{1}{3}xyz\)
\(=3xyz-\frac{1}{6}x^2y^2z^2\)
\(xyz\left(3-\frac{1}{6}xyz\right)\)
b) \(3xyz^5\cdot\left(-\frac{1}{7}\right)xyz\cdot\frac{-1}{8}xyz^4\)
\(=\left[3\cdot\left(-\frac{1}{7}\right)\cdot\left(-\frac{1}{8}\right)\right]\left(x\cdot x\cdot x\right)\left(y\cdot y\cdot y\right)\left(z^5\cdot z\cdot z^4\right)\)
\(=\frac{3}{56}x^3y^3z^{10}\)
a, \(3xyz^2+\left(\frac{-4}{8}xyz^5\right)\cdot\frac{1}{3}xyz=3xyz^2+\left[\left(\frac{-4}{8}\right)\cdot\frac{1}{3}\right]xyz^5xyz\)\(=3xyz^2-\frac{1}{2}x^2y^2z^6\)
b, \(3xyz^5\cdot\left(\frac{-1}{7}xyz^2\right)\cdot\frac{-1}{8}xyz^4=\left[3\cdot\left(\frac{-1}{7}\right)\cdot\left(\frac{-1}{8}\right)\right]xyz^5xyz^2xyz^4=\frac{3}{56}x^3y^3z^{11}\)
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
Giải :
\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)
\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)
\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)
~~Học tốt~~