\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)

\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)

\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)

\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)

\(=\left(\frac{2}{5}\right)^2\)

\(=\frac{4}{25}\)

\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)

\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)

\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)

\(=\frac{-7}{5}.\left(-10\right)\)

\(=14\)

a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)

\(=8+3.1+4:\frac{1}{2}\)

\(=8+3+8=19\)

b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)

c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)

\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)

a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)

\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)

\(=10:\left(\frac{-2}{3}\right)\)

\(=-15\)

b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\frac{-8}{27}+\frac{1}{10}\)

\(=\frac{-8}{3}+\frac{1}{10}\)

\(=\frac{-77}{30}\)

c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)

\(=\frac{-1}{3}\)

\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)

\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)

\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)

\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)

\(=-15\)

\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)

\(=-\frac{8}{3}+\frac{1}{10}\)

\(=-\frac{77}{30}\)

\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)

\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)

\(=\frac{1}{3}\)

Bài 1

\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)

\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)

\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)

\(=\frac{9}{25}+\frac{8}{9}-1\)

\(=\frac{56}{225}\)

\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)

\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)

\(=1:\frac{4}{3}=\frac{3}{4}\)

Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v 

\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)

\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)

\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)

\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)

\(=-\frac{1}{2}\)

Bài 2

\(a,\frac{2}{3}-\left(\frac{3}{4}-x\right)=\sqrt{\frac{1}{9}}\)

\(\frac{2}{3}-\left(\frac{3}{4}-x\right)=\frac{1}{3}\)

\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}\)

\(\frac{3}{4}-x=\frac{1}{3}\)

\(x=\frac{3}{4}-\frac{1}{3}\)

\(x=\frac{5}{12}\)

\(b,\left(\frac{1}{2}-x\right)^2=\left(-2\right)^2\)

\(\Rightarrow\frac{1}{2}-x=-2\) hoặc                   \(\frac{1}{2}-x=2\)

\(\Rightarrow x=\frac{1}{2}-\left(-2\right)\)                      \(x=\frac{1}{2}-2\)

\(\Rightarrow x=\frac{5}{2}\)                                         \(x=-\frac{3}{2}\)

\(c,|3x+\frac{1}{2}|-\frac{2}{3}=1\)

\(|3x+\frac{1}{2}|=1+\frac{2}{3}\)

\(|3x+\frac{1}{2}|=\frac{5}{3}\)

\(\Rightarrow\orbr{\begin{cases}3x+\frac{1}{2}=\frac{5}{3}\\3x+\frac{1}{2}=-\frac{5}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{5}{3}-\frac{1}{2}=\frac{7}{6}\\3x=-\frac{5}{3}-\frac{1}{2}=-\frac{13}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{18}\\x=-\frac{13}{18}\end{cases}}}\)

\(d,3^{2x+1}=81\)

\(\Rightarrow3^{2x+1}=3^4\)

\(\Rightarrow2x+1=4\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\frac{3}{2}\)

a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\)

b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\)

c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)

Bây giờ tạm gọi các biểu thức ở mỗi bài lần lượt là A;B;C;...

a/\(A=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}=3^2=9\)

b/\(B=\frac{3^{10}.3^5.5^5}{-5^6.3^{14}}=\frac{-3}{5}\)

c/\(C=2^3+3.1-\frac{1}{2^2}.2^2+\frac{2^2}{2}.2^3=8+3-1+16=26\)

d/\(D=\frac{3^4}{2^8}.\frac{2^{12}}{3^8}=\frac{2^4}{3^4}=\frac{16}{81}\)

e/\(E=\frac{-31^3}{2^9}.\frac{2^{20}}{31^4}=\frac{-2^{11}}{31}=\frac{-2048}{31}\)

f/\(F=\frac{-3^5}{2^{10}}.\frac{2^{20}}{3^{10}}=\frac{-2^{10}}{3^5}=\frac{-1024}{243}\)