\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{2^5.3^3.5^3.7.\left(-2\right)}{3.2^5.5^3.7}=\frac{3^2.\left(-2\right)}{1}=-18\)

\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{\left(-2\right)^3.3^3.5^3.7.2^3}{3.2^4.5^3.2.7}=\)\(\frac{\left(-2\right)^6.3^3.5^3.7}{3.2^5.5^3.7}=\frac{-2.3^2}{1}=-18\)

~~~ học tốt ~~~

-18/125

Chúc bn hok tốt!

Giải chi tiết nha m.n

\(\dfrac{-2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)

\(=\dfrac{-2^6\cdot3^3\cdot5^3\cdot7}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)

\(=\dfrac{-2^6\cdot3^3\cdot5^3}{2^5\cdot3^2\cdot5^3}=-2\cdot3=-6\)

a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

(-2).3³.5³.7.8/3.2⁴.5³.14

=(-378000)/84000

=-9/2

\(2A=2.2^3+3.2^4+4.2^5+...+100.2^{101}\)

=> \(2A-A=100.2^{101}-\left(2^{100}+2^{99}+...+2^4+2^3\right)-2.2^2\)

Đặt \(B=2^3+2^4+...+2^{100}\Rightarrow2B=2^4+2^5+...+2^{101}\)

=> \(2B-B=2^{101}-2^3\Rightarrow B=2^{101}-2^3\)

=> \(2A-A=100.2^{101}-\left(2^{101}-2^3\right)-2.2^2\)

=> \(A=\left(100.2^{101}-2^{101}\right)+2^3-2^3\)=\(99.2^{101}\)

helllo

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