Bài 1:

a) \(\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)

\(=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)

\(=\dfrac{5}{3}\)

b) \(\left(0,2\right)^2\cdot5-\dfrac{2^3\cdot27}{4^6\cdot9^5}\)

\(=0,2\cdot5\cdot0,2-\dfrac{2^3\cdot3^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=\dfrac{1}{5}-\dfrac{2^3\cdot3^3}{2^{12}\cdot3^{10}}\)

\(=\dfrac{1}{5}-\dfrac{1}{2^9\cdot3^7}\)

\(=\dfrac{2^9\cdot3^7}{2^9\cdot3^7\cdot5}-\dfrac{5}{2^9\cdot3^7\cdot5}\)

\(=\dfrac{2^9\cdot3^7-5}{2^9\cdot3^7\cdot5}\) 

c) \(\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6\cdot\left(1+2^2+2^3\right)}{26\cdot5^6}\)

\(=\dfrac{1+2^2+2^3}{26}\)

\(=\dfrac{1+4+8}{26}\)

\(=\dfrac{13}{26}\)

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

Bài 2: 

Theo đề ta có:

\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):-\dfrac{1}{4}=-\dfrac{15}{4}\)

\(\Rightarrow\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right)=-\dfrac{15}{4}\cdot-\dfrac{1}{4}\)

\(\Rightarrow a\cdot\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{15}{16}\)

\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{15}{16}-\dfrac{3}{4}\)

\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{3}{16}\)

\(\Rightarrow a=\dfrac{3}{16}:\dfrac{1}{2}\)

\(\Rightarrow a=\dfrac{3}{8}\) 

1:

a: \(=\dfrac{5^{16}\cdot3^{21}}{3^{22}\cdot5^{15}}=\dfrac{1}{3}\cdot5=\dfrac{5}{3}\)

b: \(=0.04\cdot5-\dfrac{2^3\cdot3^3}{3^6\cdot2^{12}}\)

\(=0.2-\dfrac{1}{3^3\cdot2^9}=\dfrac{1}{5}-\dfrac{1}{3^3\cdot2^9}=\dfrac{3^3\cdot2^9-5}{5\cdot3^3\cdot2^9}\)

c: \(=\dfrac{5^6+4\cdot5^6+2^3\cdot5^6}{26\cdot5^6}=\dfrac{1+4+8}{26}=\dfrac{13}{26}=\dfrac{1}{2}\)

2:

Theo đề, ta có:

\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):\dfrac{-1}{4}=\dfrac{-15}{4}\)

=>\(\dfrac{1}{2}a+\dfrac{3}{4}=\dfrac{15}{16}\)

=>1/2a=15/16-12/16=3/16

=>a=3/8

2:

a: x+201=351

=>x=351-201

=>x=150

b: \(8\cdot5^2-27:25\)

\(=8\cdot25-\dfrac{27}{25}\)

\(=200-1,08=198,92\)

d: \(2023-23:\left[9+2\left(2^3-0,21\right)\right]\)

\(=2023-23:\left[9+16-0,42\right]\)
\(=2023-\dfrac{23}{25-0,42}\)

\(=2023-\dfrac{1150}{1229}=\dfrac{2485117}{1229}\)

b: 2(x-21)=84

=>x-21=84/2=42

=>x=42+21=63

c: 135-4(81-x)=55

=>4(81-x)=135-55=80

=>81-x=20

=>x=61

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a) ( 75  - 3 2  -  12 )( 3  +  2 )

=(5 3 - 3 2 - 2 3 )( 3  +  2 )

=3( 3  -  2 )( 3  +  2 ) = 3

a)

\(\begin{array}{l}M = \frac{1}{2} + \frac{2}{3} + \left( { - \frac{1}{2}} \right) + \frac{1}{3}\\ = \frac{3}{6} + \frac{4}{6} + \left( {\frac{{ - 3}}{6}} \right) + \frac{2}{6}\\ = \frac{{3 + 4 + \left( { - 3} \right) + 2}}{6}\\ = \frac{6}{6} = 1\end{array}\)

b)

\(\begin{array}{l}M = \frac{1}{2} + \frac{2}{3} + \left( { - \frac{1}{2}} \right) + \frac{1}{3}\\ = \left[ {\frac{1}{2} + \left( {\frac{{ - 1}}{2}} \right)} \right] + \left[ {\frac{2}{3} + \frac{1}{3}} \right]\\ = 0 + 1 = 1\end{array}\)

(3 - 2i)(2 - 3i) = 3. 2 – 3. 3i - 2i.2 - 2i. (- 3i) = 6 – 9i – 4i – 6 = ( 6- 6) + ( -9 – 4).i = -13i

Ta có: (3 - 5i) + (2 + 4i) = (3 + 2) + (-5 + 4)i = 5 - i

Lời giải:

$10+2.4^2=10+2.16=10+32=42$

Đáp án A

$50-4.3^2=50-4.9=50-36=14$

Đáp án A.