\(N=\dfrac{3^{10}.11+3^9.15}{3^9.2^4}=\dfrac{3^9.33+3^9.15}{3^9.2^4}\)

\(=\dfrac{3^9\left(33+15\right)}{3^9.16}\)

\(=\dfrac{48}{16}=3\)

\(A=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\cdot\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}\cdot16}{3^9\cdot16}=3\)

\(B=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\cdot\left(13+65\right)}{2^8\cdot2^2\cdot26}=\dfrac{2^{10}\cdot78}{2^{10}\cdot26}=3\)

\(C=\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(3^3\cdot2^2\right)^4}\\ =\dfrac{2^9\cdot3^6\cdot2^4\cdot3^6}{3^{12}\cdot2^8}=\dfrac{2^{13}\cdot3^{12}}{3^{12}\cdot2^8}=2^5=32\)

\(D=\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}\\ =\dfrac{3^{29}\cdot\left(11-3\right)}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{4\cdot3^{28}}=3\cdot2=6\)

may mà m đăng để t đỡ phải viết (mỏi tay) Nguyen Ngoc Linh

\(A=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+2^{12}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^{19}+2^{12}}=\dfrac{2^{10}+5.2^8}{3^{10}+2^3}=\dfrac{2^7+5.2^5}{3^{10}}\)

\(B=\dfrac{4}{35}+\dfrac{4}{63}+\dfrac{4}{99}+\dfrac{4}{143}+\dfrac{4}{195}\)

\(=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\right)\)

\(=2.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{15}\right)=2\left(\dfrac{2}{15}\right)\)=\(\dfrac{4}{15}\)

Lời giải:
a.

\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)

\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)

b.

\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 10B\Rightarrow A< B$

Ko cần biet vi ko biet ang ang

\(\dfrac{1}{2022}\) \(\times\) \(\dfrac{2}{5}\) + \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{7}{5}\) - \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{8}{10}\)

= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{2}{5}\) + \(\dfrac{7}{5}\) - \(\dfrac{8}{10}\))

= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{9}{5}\) - \(\dfrac{4}{5}\))

= \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{5}{5}\)

=  \(\dfrac{1}{2022}\times1\)

= \(\dfrac{1}{2022}\)

Q=(2^9.3+2^9.5):2^12

Đặt A=2^9.3+2^9.5

      A=2^9.(3+5)

      A=2^9.8

Mặt khác:8=2^3

=>A=2^9.2^3

     A=2^12

Theo đề bài ta có Q=(2^9.3+2^9.5):2^12

=>Q=2^12:2^12

    Q=1

Nhìn dài dòng thế thôi chứ đơn giản lắm.Nếu thấy đúng thì cho mình nhé!

`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`

`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`

`=sqrt{10}-(2+sqrt{10})`

`=-2`

`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`

`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`

`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`

`=(5sqrt5-5sqrt3-4)/2`

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}+\frac{2^{13}+2^5}{2^{10}+2^2}=11\)