\(=\left(4\times2^5\right):\left(\frac{8.1}{16}\right)=128:\frac{1}{2}=256\)

a) \(25^3:5^2=5^6:5^2=5^4=625\)

b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)

c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\frac{1}{8}=\frac{17}{8}\)

a) \(25^3:5^2=5^6:5^2=5^{6-2}=5^4\)

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

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

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

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)

\(=\dfrac{3}{8}+\dfrac{5}{8}\)

\(=1\)

toán lớp 7

bn đổi ngược hai vế cho nhau là ra 1 bài toán bình thường thôi

Bài nỳ tuy rất dài nhưng cũng dễ

Chí cần cậu cuyển VT sang VP rồi tìm x bình thường

Chúc cậu học tốt

\(\left(0,66-0,012:0,2\right):\left(1-1\frac{4}{4}\times0,4\right)=2\frac{9}{13}.x:\left(3,125-5,6:2\frac{2}{3}\right)\)

\(2\frac{9}{13}.x:\frac{41}{40}=\frac{3}{5}:\frac{13}{35}\)

\(2\frac{9}{13}.x:\frac{41}{40}=\frac{21}{13}\)

\(2\frac{9}{13}.x=\frac{861}{520}\)

\(x=\frac{123}{200}\)

b, \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)

Giống: Cả hai đều có số giống nhau 

Khác: Một bên là chia 2 nhân 3

          Một bên là chia 3 nhân 2

Giống: các số đều giống nhau 

Khác: vế 1 là : 2 x 3 còn vế 2 là :3x2

K mk nha, mk k lại cho

ko đón tiếp mời đi cho tui k lần 1 rùi ko có lần 2 đâu..........

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

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

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

chịu thôi

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\(A=\frac{1.25.10:0,25.24,4.2}{6,1.2.6,25.4:0,5}.\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)

\(A=\frac{50.48,8}{6,1.100}\times\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{18}{19}\times\frac{19}{20}\)

\(A=\frac{4}{1}\times\frac{1}{1}\times\frac{1}{20}\)

\(A=\frac{4}{20}\)