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Bài đầu đơn giản rồi , tự tính nhé <3
Bài 2
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.3^2-2^n.2^2+3^n-2^n\)
\(=\left(3^n.3^2+1\right)-\left(2^n.2^2+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\)
Vậy.....
\(\left(\frac{4}{9}+\frac{1}{3}\right)^2=\left(\frac{4}{9}+\frac{3}{9}\right)^2=\left(\frac{7}{9}\right)^2=\frac{49}{81}\)
\(\left(\frac{1}{2}-\frac{3}{5}\right)^3=\left(\frac{5}{10}-\frac{6}{10}\right)^3=\left(\frac{-1}{10}\right)^3=\frac{-1}{1000}\)
\(\left(\frac{-1}{5}\right)^5.\left(\frac{-6}{5}\right)^4=\frac{-5}{3125}.\frac{1296}{625}=\frac{-1296}{390625}\)
\(\left(\frac{3}{4}\right)^3:\left(\frac{3}{4}\right)^2:\left(-\frac{2}{5}\right)^3=\frac{3}{4}:\frac{-8}{125}=\frac{3}{4}.\frac{-125}{8}=\frac{-375}{32}\)
.
a) (4/9 + 1/3)^2
= (4/9 + 3/9)^2
= (7/9)^2
= 49/81
b) (1/2 - 3/5)^3
= (5/10 - 6/10)^3
= (-1/10)^3
= -1/1000
c) (-10/3)^5 x (-6/5)^4
= tương tự những câu trên
d) cũng tương tự nhé
a, \(\left(\frac{4}{9}+\frac{1}{3}\right)^2=\frac{49}{81}\)
b, \(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4=\frac{-2560}{3}\)
c, \(\left(\frac{1}{2}-\frac{3}{5}\right)^3=\frac{-1}{1000}\)
d,\(\left(\frac{3}{4}\right)^3:\left(\frac{3}{4}\right)^2:\left(\frac{-2}{3}\right)^3=\frac{-81}{32}\)
\(A=\left(\frac{3}{4}\right)^{-4}.\left(\frac{-2}{3}\right)^{-3}\)
\(A=\frac{256}{81}.\frac{-27}{8}\)
\(A=\frac{729}{64}\)
\(B=\left(4^3\right)^{-2}.a^{2015}\)
\(B=64^{-2}.a^{2015}\)
\(B=\frac{1}{4096}.a^{2015}\)
\(C=\left[\left(\frac{-1}{3}\right).\frac{2}{5}.\left(\frac{-3}{4}\right)\right]^3\)
\(C=\left[\frac{1}{10}\right]^3\)
\(C=\frac{1}{1000}\)
A =\(-\frac{32}{3}\)
B = \(\frac{1}{4096}.a^{2015}\)
C =\(\frac{1}{1000}\)