A = \(\frac{6x\left(1+8+27\right)}{6x\left(12+96+324\right)}\)= \(\frac{36}{432}\)=\(\frac{1}{12}\)

A= 1*2*3=2*4(6+3*6*9/2*3*12+4*6*24+6*9*36 = 8843366808

đề vớ vẩn

\((\frac{1}{2}+\frac{1}{4}+\frac{1}{6}) \)x \((\frac{36}{9}-\frac{12}{9}:\frac{1}{3}) \)

\(= (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}) \) x \((\frac{36}{9}-\frac{36}{9})\)

\( = (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}) \) x \(0\)

\(= 0\)

Tham khảo :

^{(Tại mik lười ko mún ghi nên bạn tham khảo cái ảnh này nhá)}

Tham khảo :

Mình vừa học xong 

a) Ta có:

\(8^9+7^9+6^9+...+1^9\)

\(=\left(8^3+7^3+6^3+...+1^3\right)^2\)

\(=\left(\left(8+7+6+...+2+1\right)^2\right)^2\)

\(=\left(8+7+6+...+2+1\right)^4\)

\(=36^4=9^4.4^4\)

Mà \(9^{10}=9^4.9^6\)

\(\Rightarrow9^4.9^6>9^4.4^4\)

Vậy \(9^{10}>8^9+7^9+6^9+...+1^9\)

b) \(45=5.9\)

Ta có:

\(\left\{{}\begin{matrix}36⋮9\\9⋮9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}36^{36}⋮9\\9^{10}⋮9\end{matrix}\right.\)\(\Rightarrow\left(36^{36}-9^{10}\right)⋮9\)

Lại có:

\(36\div5\) dư \(1\)

\(9\div5\) dư \(1\)

\(\Rightarrow\left(36^{36}-9^{10}\right)⋮5\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) và \(\left(9;5\right)=1\)

\(\Rightarrow\left(36^{36}-9^{10}\right)⋮45\) (Đpcm)

=(1/2*6)+(1/38*60)

=3+30/19=87/19

\(\dfrac{1}{3}\cdot\dfrac{-6}{13}\cdot\dfrac{-9}{10}\cdot\dfrac{-13}{36}=\dfrac{1}{3}\cdot\dfrac{-6}{13}\dfrac{-9}{10}\dfrac{-1}{2}=\dfrac{1\cdot\left(-6\right)\cdot\left(-9\right)\left(-1\right)}{3\cdot13\cdot10\cdot2}=\dfrac{-1\cdot2\cdot3\cdot-9}{3\cdot13\cdot10\cdot2}=\dfrac{9}{130}\)