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a,\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{13}{14}\right)^2\)
\(=\dfrac{169}{196}\)
b,\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=\left(\dfrac{-1}{12}\right)^2\)
\(=\dfrac{1}{144}\)
c,\(\dfrac{5^4.20^4}{25^5.4^5}\)
\(=\dfrac{100^4}{100^5}\)
\(=\dfrac{1}{100}\)
d,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)
\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=4^4.\left(\dfrac{-10}{3}\right)\)
\(=256.\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{-2560}{3}\)
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
a) \(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\frac{1}{3^4}.3^2=3^7.\frac{1}{3^4}=3^3\)
b) \(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{16}=2^7:2^3.16=2^4.2^4=2^8\)
c) \(3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=2^5.2^2=2^7\)
d) \(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^3.\left(3^2\right)^2=\frac{1^3}{3^3}.3^4=1^3.3=3^1\)
a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
a) \(12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12\cdot\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{12\cdot4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{16+4}{3}\)
\(=\dfrac{20}{3}\)
b) \(\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}\cdot\left(-\dfrac{1}{2}\right)^2\right]\)
\(=\dfrac{9}{4}-\left(\dfrac{1}{2}:2-9\cdot\dfrac{1}{4}\right)\)
\(=\dfrac{9}{4}-\left(\dfrac{1}{4}-9\cdot\dfrac{1}{4}\right)\)
\(=\dfrac{9}{4}-\dfrac{1}{4}\cdot\left(1-9\right)\)
\(=\dfrac{9}{4}+\dfrac{8}{4}\)
\(=\dfrac{17}{4}\)
c) \(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\)
\(=-\dfrac{1}{12}:\dfrac{5}{11}+\dfrac{1}{12}\)
\(=\dfrac{1}{12}\cdot-\dfrac{11}{5}+\dfrac{1}{12}\)
\(=\dfrac{1}{12}\cdot\left(-\dfrac{11}{5}+1\right)\)
\(=\dfrac{1}{12}\cdot-\dfrac{6}{5}\)
\(=-\dfrac{1}{10}\)
d) \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)
\(=-\dfrac{1}{15}+\dfrac{4}{9}:\left(2+\dfrac{2}{3}\right)-\dfrac{5}{6}\)
\(=-\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)
\(=-\dfrac{9}{10}+\dfrac{1}{6}\)
\(=-\dfrac{11}{15}\)
e) \(\dfrac{3^7\cdot8^6}{6^6\cdot\left(-2\right)^{12}}\)
\(=\dfrac{3^7\cdot\left(2^3\right)^6}{2^6\cdot3^6\cdot2^{12}}\)
\(=\dfrac{3^7\cdot2^{18}}{2^{6+12}\cdot3^6}\)
\(=\dfrac{2^{18}\cdot3^7}{2^{18}\cdot3^6}\)
\(=3^{7-6}\)
\(=3\)
\(a,12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\\ =12\cdot\dfrac{4}{9}+\dfrac{4}{3}\\ =\dfrac{16}{3}+\dfrac{4}{3}\\ =\dfrac{20}{3}\\ b,\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}.\left(-\dfrac{1}{2}\right)^2\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{2}\cdot\dfrac{1}{2}-9\cdot\dfrac{1}{4}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{8}{4}\right)\\ =\dfrac{17}{4}\)
\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =\left(-\dfrac{9}{12}+\dfrac{8}{12}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{3}{12}+\dfrac{4}{12}\right)\\ =-\dfrac{1}{12}\cdot\dfrac{11}{5}+\dfrac{1}{12}\\ =-\dfrac{11}{60}+\dfrac{1}{12}\\ =-\dfrac{1}{10}\)
\(d,\dfrac{-1^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left(-\dfrac{5}{6}\right)\\ =-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}+\dfrac{5}{6}\\ =-\dfrac{1}{15}+\dfrac{1}{6}+\dfrac{5}{6}\\ =\dfrac{1}{10}+\dfrac{5}{6}\\ =\dfrac{14}{15}\)
`e,` Không hiểu đề á c: )
a.\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
=\(\left(\frac{6}{14}+\frac{7}{14}\right)^2\)
=\(\left(\frac{13}{14}\right)^2\)
=\(\frac{13^2}{14^2}\)
=\(\frac{169}{196}\)
b.\(\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
=\(\left(\frac{9}{12}-\frac{10}{12}\right)^2\)
=\(\left(\frac{-1}{12}\right)^2\)
=\(\frac{-1^2}{12^2}\)
=\(\frac{1}{144}\).
c.Phần C bn viết lại đề bài đi,mk ko hiểu
d.\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)
=\(\frac{-10^5}{3^5}.\left(\frac{-6^4}{5^4}\right)\)
=\(\frac{-100000}{243}.\frac{1296}{625}\)
=\(\frac{-2560}{3}\)
Không biết đúng ko nữa
c.\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{25^2.\left(5.4\right)^4}{25^5.4^5}\)
=\(\frac{1.\left(5\right)^4.\left(4\right)^4}{25^3.4^5}\)
=\(\frac{25^2.1}{25^3.4}\)
=\(\frac{1}{25.4}\)
=\(\frac{1}{100}\).