21158

\(\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)

mấy cái kia tương tự mình sẽ giải cho bạn cái cuối cùng:

\(\dfrac{\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^{n-1}}=\dfrac{\left(-\dfrac{1}{2}\right)^{n-1}}{\left(-\dfrac{1}{2}\right)^{n-1}}.\left(-\dfrac{1}{2}\right)=1.\left(-\dfrac{1}{2}\right)=-\dfrac{1}{2}\)

\(\dfrac{5}{4x13}\)+\(\dfrac{5x7}{4x13}\)+ \(\dfrac{10}{13}\)

=\(\dfrac{5}{4x13}\)+ \(\dfrac{5}{13}\)x(\(\dfrac{7}{4}\)+2)

=\(\dfrac{5}{13}\)(\(\dfrac{1}{4}\)+\(\dfrac{7}{4}\)+2)

=\(\dfrac{5}{13}\)x4

=\(\dfrac{20}{13}\)

+) \(2^x.162=1024\)

\(\Leftrightarrow2^x=\dfrac{512}{81}\approx6,3\)

\(\Rightarrow x\approx2,66\left(ktm\right)\)

=> không tìm được x thỏa mãn 

+) \(64.4^x=168\)

\(\Leftrightarrow4^x=2,625\)

\(\Leftrightarrow x\approx0,7\)

=> không tìm được x thỏa mãn 

Ta có: `\hat{xOy}+\hat{xOz}=180^o` (`2` góc kề bù)

  `=>120^o +\hat{xOz}=180^o`

  `=>\hat{xOz}=60^o`

Vì `Om` là tia p/g của `\hat{xOy}=>\hat{xOm}=1/2\hat{xOy}=1/2 .120^o=60^o`

Ta có: `\hat{xOm}+\hat{xOz}=\hat{mOz}`

   `=>\hat{mOz}=60^o +60^o =120^o`

Vì Om là phân giác của \(\widehat{mOz}\)

\(\Rightarrow\widehat{xOm}=\dfrac{1}{2}\widehat{xOy}=\dfrac{1}{2}.120^o=60^o\)

Vì \(\widehat{xOy}\) và \(\widehat{xOz}\) là 2 góc kề bù 

\(\Rightarrow\widehat{xOz}=180^o-120^o=60^o\) 

\(\Rightarrow\widehat{mOz}=\widehat{xOz}+\widehat{xOm}=60^0.2=120^o\)

\(\dfrac{3^{12}+3^{15}}{1+3^3}\)

\(=\dfrac{3^{12}\left(1+3^3\right)}{1+3^3}\)

\(=3^{12}\)

\(\dfrac{3^{12}+3^{15}}{1+3^3}=\dfrac{3^{12}\left(1+3^3\right)}{1+3^3}=3^{12}\)

\(\dfrac{-2}{3}=\dfrac{7}{x}\)

\(=>-2x=3.7=21\)

\(=>x=\dfrac{-21}{2}\)