\(A=\dfrac{\left(-3\right)^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)^{45}\cdot\left(-2\right)^{12}}{5\cdot\left(-3\right)^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)\cdot\left(-2\right)}{5}=\dfrac{6}{5}\)

Thank u very much!!

Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)

\(\dfrac{4^3\cdot9^5\cdot\left(-2\right)^6}{16^4\cdot\left(-27\right)^2}\)

=\(\dfrac{\left(2^2\right)^3\cdot\left(3^2\right)^5\cdot2^6}{\left(2^4\right)^4\cdot\left(-3^3\right)^2}\)

=​​​\(\dfrac{2^6\cdot3^{10}\cdot2^6}{2^{16}\cdot\left(-3\right)^6}\)​

=\(\dfrac{2^{6+6}\cdot3^{10}}{2^{16}\cdot3^6}\)

=\(\dfrac{2^{12}\cdot3^{10}}{2^{16}\cdot3^6}\)

=\(\dfrac{3^4}{2^4}\)

=\(\dfrac{81}{16}\)

Lời giải:
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{(3^2)^4.(3^3)^5.3^6.3^4}{3^8.(3^4)^4.2.3^2.13.(2^3)^2}\)

\(=\frac{3^8.3^{15}.3^6.3^4}{3^8.3^{16}.2.3^2.13.2^6}=\frac{3^{33}}{3^{26}.2^7.13}=\frac{3^7}{2^7.13}\)

\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)

\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)

Bài 1:

a: \(4\cdot64\cdot2^8\)

\(=2^2\cdot2^6\cdot2^8\)

\(=2^{2+6+8}=2^{16}\)

b: \(128\cdot27=2^7\cdot3^3\)

c: \(4\cdot27:\left(3^{11}\cdot\frac19\right)=4\cdot27:3^9=4\cdot\frac{3^3}{3^9}=\frac{4}{3^6}=\left(\frac{2}{3^3}\right)^2\)

Bài 2:

a: \(\left(\frac12\right)^3\cdot4+\frac34\)

\(=4\cdot\frac18+\frac34\)

\(=\frac48+\frac34=\frac48+\frac68=\frac{10}{8}=\frac54\)

b: Sửa đề: \(4^6\cdot\left(\frac12\right)^{12}\)

Ta có: \(4^6\cdot\left(\frac12\right)^{12}\)

\(=\left(2^2\right)^6\cdot\frac{1}{2^{12}}\)

\(=\frac{2^{12}}{2^{12}}=1\)

c: Ta có: \(\left(\frac12\right)^5-1,5^2\)

\(=\frac{1}{32}-\frac94\)

\(=\frac{1}{32}-\frac{72}{32}=\frac{-71}{32}\)

d: \(\frac{14^{16}\cdot35^7}{10^9\cdot7^{22}}\)

\(=\frac{7^{16}\cdot2^{16}\cdot7^7\cdot5^7}{2^9\cdot5^9\cdot7^{22}}\)

\(=\frac{7^{23}}{7^{22}}\cdot\frac{2^{16}}{2^9}\cdot\frac{5^7}{5^9}=7\cdot\frac{2^7}{5^2}=7\cdot\frac{128}{25}=\frac{896}{25}\)

ĐKXĐ : \(x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}\ne0\)

\(\Leftrightarrow x\ne\sqrt[4]{2}\)

\(P=\dfrac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}}\)

\(=\dfrac{x^2-\sqrt{2}}{\left(x^4-\sqrt{2}x^2\right)+\sqrt{3}\left(x^2-\sqrt{2}\right)}\) 

\(=\dfrac{x^2-\sqrt{2}}{\left(x^2+\sqrt{3}\right)\left(x^2-\sqrt{2}\right)}=\dfrac{1}{x^2+\sqrt{3}}\)

`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`