Biểu thức 1

$3\dfrac12\cdot\dfrac4{49}-\left(2,(4)\cdot2\dfrac5{11}\right):\left(-\dfrac{42}{8}\right)$

$=\dfrac72\cdot\dfrac4{49}-\left(\dfrac{22}{9}\cdot\dfrac{27}{11}\right)\cdot\left(-\dfrac4{21}\right)$

$=\dfrac27-6\cdot\left(-\dfrac4{21}\right)$

$=\dfrac27+\dfrac87$

$=\dfrac{10}{7}.$

Biểu thức 2

$\left[0,(5)\cdot0,(2)\right]:\left(3\dfrac13:\dfrac{33}{25}\right)-\left(\dfrac25\cdot3\dfrac13\right):\dfrac43$

$=\left(\dfrac59\cdot\dfrac29\right):\left(\dfrac{10}{3}:\dfrac{33}{25}\right)-\left(\dfrac25\cdot\dfrac{10}{3}\right):\dfrac43$

$=\dfrac{10}{81}:\dfrac{250}{99}-\dfrac43\cdot\dfrac34$

$=\dfrac{10}{81}\cdot\dfrac{99}{250}-1$

$=\dfrac{11}{225}-1$

$=-\dfrac{214}{225}.$

Biểu thức 1

$A=3\dfrac12\cdot\dfrac4{49}\cdot\left(2,(4)\cdot2\dfrac5{11}\right):\left(-\dfrac{42}{5}\right)$

$=\dfrac72\cdot\dfrac4{49}\cdot\left(\dfrac{22}{9}\cdot\dfrac{27}{11}\right)\cdot\dfrac5{-42}$

$=\dfrac72\cdot\dfrac4{49}\cdot6\cdot\dfrac5{-42}$

$=\dfrac{12}{7}\cdot\dfrac5{-42}$

$=\dfrac{60}{-294}$

$=-\dfrac{10}{49}.$

Biểu thức 2

$B=\left[0,(5)\cdot0,(2)\right]:\left(3\dfrac13:\dfrac{33}{25}\right)-\left(\dfrac25\cdot1\dfrac13\right):\dfrac43$

$=\left(\dfrac59\cdot\dfrac29\right):\left(\dfrac{10}{3}:\dfrac{33}{25}\right)-\left(\dfrac25\cdot\dfrac43\right):\dfrac43$

$=\dfrac{10}{81}:\dfrac{250}{99}-\dfrac{8}{15}\cdot\dfrac34$

$=\dfrac{10}{81}\cdot\dfrac{99}{250}-\dfrac25$

$=\dfrac{11}{225}-\dfrac{90}{225}$

$=-\dfrac{79}{225}.$

Câu 1:

(3/5)^15 : (9/25)^5

= (3/5)^15 : (3/5)^10

= (3/5)^5

Câu 2:

5 - (-5/11)^0 + (1/3)^2 - 3

= 5 - 1 + 1/9 - 3

= (5 - 1 - 3) + 1/9

= 1 + 1/9

= 10/9

a, 27³ : 3² = (3³)³ : 3² = 3⁹ : 3² = 3^9-2 = 3⁷b, 

Câu a:

27^3 : 3^2

= 3^9 : 3^2

= 3^7

\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)

\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)

\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)

C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)

C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)

C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)

C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)

C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)

C = \(\frac{-233}{135}\)

D =  \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)

D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)

D = \(-4.\frac{12}{13}\)

D = \(\frac{-48}{13}\)

E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)

E = \(5.4-4.3+5-0,3.20\)

E = \(20-12+5-6\)

E = \(8+\left(-1\right)\)

E = \(7\)

F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)

F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)

F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\) 

F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)

F = \(\frac{-11}{12}\)

 Chúc cậu hk tốt ~ 

$\textbf{a)}$

$4\cdot\left(\dfrac14\right)^2+25\cdot\left[\left(\dfrac34\right)^3:\left(\dfrac54\right)^3\right]:\left(\dfrac32\right)^3$

$=\dfrac14+25\cdot\left(\dfrac35\right)^3:\left(\dfrac32\right)^3$

$=\dfrac14+25\cdot\dfrac{27}{125}\cdot\dfrac8{27}$

$=\dfrac14+\dfrac85$

$=\dfrac{5+32}{20}$

$=\boxed{\dfrac{37}{20}}.$

$\textbf{b)}$

$2^3+3\cdot\left(\dfrac12\right)^0+(-1)+\left[(-2)^2:\dfrac12\right]-8$

$=8+3\cdot1-1+\dfrac4{1/2}-8$

$=8+3-1+8-8$

$=\boxed{10}.$

$\textbf{A}$

$A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}$

$=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+2^3\cdot5^9\cdot7^3}$

$=\dfrac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\dfrac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+8)}$

$=\dfrac{2}{12}-\dfrac{-30}{9}$

$=\dfrac16+\dfrac{10}{3}$

$=\boxed{\dfrac72}.$

$\textbf{B}$

$B=\dfrac{\left(-\dfrac12\right)^3-\left(\dfrac34\right)^3\cdot(-2)^2}{2\cdot(-1)^5+\left(\dfrac34\right)^2-\dfrac38}$

$=\dfrac{-\dfrac18-\dfrac{27}{64}\cdot4}{-2+\dfrac9{16}-\dfrac38}$

$=\dfrac{-\dfrac18-\dfrac{27}{16}}{-2+\dfrac3{16}}$

$=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}$

$=\boxed{1}.$

$\textbf{a)}$

$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$

$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$

$=-\dfrac{13}{8}+\dfrac{39}{56}$

$=-\dfrac{13}{14}.$

$\textbf{b)}$

$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$

$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$

$=\dfrac19.$