\(\frac{35\left(27^8+2.9^{11}\right)}{15\left(81^6-12.3^{19}\right)}\)= \(\frac{35\left(\left(3^3\right)^8+2.\left(3^2\right)^{11}\right)}{15\left(\left(3^4\right)^6-4.3.3^{19}\right)}\)= \(\frac{35\left(3^{24}+2.3^{22}\right)}{15\left(3^{24}-4.3^{20}\right)}\)= \(\frac{35\left(3^{22}.3^2+2.3^{22}\right)}{15\left(3^{20}.3^4-4.3^{20}\right)}\)= \(\frac{35\left(3^{22}.\left(9+2\right)\right)}{15\left(3^{20}.\left(81-4\right)\right)}\)= \(\frac{35\left(3^{22}.11\right)}{15\left(3^{20}.77\right)}\)= \(\frac{5.7.3^{22}.11}{5.3.3^{20}.7.11}\)= \(\frac{3^{22}}{3.3^{20}}\)= \(\frac{3^{20}.3.3}{3.3^{20}}\)= \(\frac{3}{1}\)= 3

Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak

Bài 2:

\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)

\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)

\(=\dfrac{1}{313}\)

\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)

\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)

\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)

\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)

\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)

\(=\dfrac{7.2.10}{81-4}\)

\(=\dfrac{720}{77}\)

@Phùng Khánh Linh, Nguyễn Nam, Ribi Nkok Ngok, Trần Quốc Lộc, Phạm Hoàng Giang, Thien Tu Borum, Nguyễn Thanh Hằng, Thảo Phương , Trương Hồng Hạnh, ...

Bài 1:
\(\frac{35(27^8+2.9^{11})}{15(81^6-12.3^{19})}=\frac{5.7(3^{24}+2.3^{22})}{3.5(3^{24}-2^2.3^{20})}\\ =\frac{5.7.3^{22}(3^2+2)}{3.5.3^{20}(3^4-2^2)}\\ =\frac{5.7.3^{22}.7}{3.5.3^{20}.7.11}\\ =\frac{7.3}{11}=\frac{21}{11}\)

Bài 2:

a. $(2x+1)(y-5)=10$

Với $x,y$ tự nhiên thì $2x+1$ là số tự nhiên lẻ và $y-5$ là số nguyên.

Mà tích của chúng bằng $10$ nên ta xét các TH sau:

TH1: $2x+1=1, y-5=10\Rightarrow x=0; y=15$

TH2: $2x+1=5, y-5=2\Rightarrow x=2; y=7$

b.

$x(y+2)-y=5$

$x(y+2)-(y+2)=3$

$(x-1)(y+2)=3$
Với $x,y$ tự nhiên thì $y+2$ là số tự nhiên, $x-1$ là số nguyên. Mà tích của chúng bằng $3$ nên ta xét các TH sau:

TH1: 

$y+2=1, x-1=3\Rightarrow y=-1, x=4$ (loại vì $y=-1$ không là stn)

TH2: 

$y+2=3, x-1=1\Rightarrow y=1, x=2$

Bài 3:

a.

$3^{21}=3.3^{20}=3.9^{10}>2.8^{10}=2.2^{30}=2^{31}$

b.

$63^7< 64^7=(4^3)^7=4^{21}< 4^{24}=(4^2)^{12}=16^{12}$