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b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
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: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
\(a,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\left(\frac{3}{7}\right)^2\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6=\left(\frac{9}{49}\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6\)
\(=\left(\left(\frac{9}{49}\right)^{10}:\left(\frac{9}{49}\right)^6\right).\frac{3}{7}=\left(\frac{9}{49}\right)^{10-6}.\frac{3}{7}=\left(\frac{9}{49}\right)^4.\frac{3}{7}=\left(\left(\frac{3}{7}\right)^2\right)^4.\frac{3}{7}\)
\(=\left(\frac{3}{2}\right)^8.\frac{3}{7}=\left(\frac{3}{2}\right)^9\)
\(b,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}=2+\left(\frac{1}{2}\right)^3=2+\frac{1}{6}=2\frac{1}{6}\)
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,
\(\left(-\dfrac{1}{3}\cdot x\right)^2\cdot y\cdot2\cdot xy^3 \)
\(=\left(-\dfrac{1}{3}\right)^2\cdot x^2y\cdot2xy^3\)
\(=\left(-\dfrac{1}{9}\right)\cdot2\cdot\left(x^2\cdot x\right)\cdot\left(y\cdot y^3\right)\)
\(=-\dfrac{2}{9}\cdot x^3\cdot y^4\)
b,\(=\left(\dfrac{1}{4}\cdot x\right)^2.\left(-2\right)\cdot x^3y^5\)
=\(\dfrac{1}{4}^2\cdot x^2y\cdot\left(-2\right)\cdot x^3y^5\)
=\(\dfrac{1}{16}\cdot\left(-2\right)\cdot\left(x^3\cdot x^2\right)\cdot\left(y\cdot y^5\right)\)
=\(-\dfrac{1}{8}\cdot x^5y^6\)
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,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)
\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=4^4.\left(\dfrac{-10}{3}\right)\)
\(=256.\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{-2560}{3}\)
Ta có: \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right).n.\left(n+1\right)}=\frac{1}{2}\left[\frac{1}{\left(n-1\right).n}-\frac{1}{n\left(n+1\right)}\right]\)
Áp dụng ta được:
\(A< \frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{5.6}+\frac{1}{5.6}-\frac{1}{6.7}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{4.5}-\frac{1}{20.21}\right)\)
\(< \frac{1}{2}.\frac{1}{4.5}=\frac{1}{40}\)