Bài tập bổ sung 2 : Tính :
a. ( 1 + 2/3 - 1/4 ) . ( 4/5 - 3/4 ) mũ 2
b. 2 : ( 1/2 - 2/3 ) mũ 3
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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,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)
\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=4^4.\left(\dfrac{-10}{3}\right)\)
\(=256.\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{-2560}{3}\)
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
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) \(\frac{16}{2^n}=2\)
=> 2.2n = 16
=> 21+n = 24
=> 1 + n = 4
=> n = 4 - 1
=> n = 3
Vậy n = 3
b) \(\frac{\left(-3\right)^n}{81}=-27\)
=> (-3)n = -27.81
=> (-3)n = -33.34
=> (-3)n = (-3)7
=> n = 7
Vậy n = 7
c) 8n : 2n = 4
=> (8 : 2)n = 4
=> 4n = 41
=> n = 1
Vậy n = 1
a: \(\Leftrightarrow2^5\ge2^n>2^2\)
=>2<n<=5
hay \(n\in\left\{3;4;5\right\}\)
b: \(\Leftrightarrow3^2\cdot3^3\le3^n\le3^5\)
=>5<=n<=5
=>n=5
Ta có công thức tổng quát như sau:
\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)
Áp dụng ta có:
\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\)
\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)
______
\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)
\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)
_____
\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)
\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)
a.\(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
=\(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
=\(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
=\(\frac{17}{12}.\frac{1}{400}\).
=\(\frac{17}{4800}\)
b.\(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3\)
=\(2:\left(\frac{3}{6}-\frac{4}{6}\right)^3\)
=\(2:\left(\frac{-1}{6}\right)^3\)
=\(2:\left(\frac{-1}{216}\right)\)
=\(\frac{-216.2}{1}\)
=-432.
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