Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 = (1 - 2/3 + 4/3) - (4/5 - 1) + (7/5 + 2)
A= (3/3 - 2/3 + 4/3) - (4/5 - 5/5) + (7/5 + 10/5)
A= 5/3 + 1/5 + 17/5
A= 5/3 +18/5
A= 25/15 + 54/15
A= 79/15
B= (-3 + 3/4 - 1/3 ) : (5 + 2/5 - 2/3)
B= (-36/12 + 9/12 - 4/12) : (75/15 + 6/15 - 10/15)
B= -31/12 : 71/15
B= -155/284
C= (3/5 - 4/5 ) . (2/7 - 3/14) - (5/9 - 7/27) . (1 - 3/5) + (1 - 11/12) . (1-11/12)
C= -1/5 . 1/14 - 8/27 . 2/5 + 1/12 . 1/12
C=-1/70 - 16/135 + 1/144
C=-216/15120 - 1792/15120 + 105/15120
C=-1903/15120
\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)
\(P=\frac{37}{60}\)
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)
\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)
\(Q=\frac{1044}{4375}\)
\(A=\left(\frac{3}{4}\right)^{-4}.\left(\frac{-2}{3}\right)^{-3}\)
\(A=\frac{256}{81}.\frac{-27}{8}\)
\(A=\frac{729}{64}\)
\(B=\left(4^3\right)^{-2}.a^{2015}\)
\(B=64^{-2}.a^{2015}\)
\(B=\frac{1}{4096}.a^{2015}\)
\(C=\left[\left(\frac{-1}{3}\right).\frac{2}{5}.\left(\frac{-3}{4}\right)\right]^3\)
\(C=\left[\frac{1}{10}\right]^3\)
\(C=\frac{1}{1000}\)
A =\(-\frac{32}{3}\)
B = \(\frac{1}{4096}.a^{2015}\)
C =\(\frac{1}{1000}\)