(-7/2)^2222×(-2/999)^1234
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12 tháng 8 2020
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)
12 tháng 8 2020
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2A=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(2A-A=A\)
\(=\left(3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2012}}\)
\(=2-\frac{1}{2012^2}\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{6}{12}-\frac{4}{12}-\frac{2}{12}\right)\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot0=0\)
18 tháng 8 2023
\(=\sqrt{4\sqrt{3}+2\left(2-\sqrt{3}\right)}\)
\(=\sqrt{4\sqrt{3}+4-2\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
7 tháng 3 2022
Bài 4:
b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)
c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)
NP
0
NT
4
13 tháng 7 2019
5(2x - 3)(2x + 3) - 6(x - 7)2
= 5(2x - 3)(2x + 3) - 6(x2 - 14x + 49)
= 5(4x2 - 9) - 6(x2 - 14x + 49)
= 5.4x2 + 5.(-9) + (-6).x2 + (-6).(-14x) + (-6).49
= 20x2 - 45 - 6x2 + 84x - 294
= 14x2 + 84x - 339
A = (\(\dfrac{-7}{2}\))2222 \(\times\) (- \(\dfrac{2}{999}\))1234
A = \(\dfrac{7^{2222}}{2^{2222}}\) \(\times\) \(\dfrac{2^{1234}}{9^{1234}}\)
A = \(\dfrac{7^{2222}}{9^{1234}}\) \(\times\) \(\dfrac{2^{1234}}{2^{2222}}\)
A = \(\dfrac{7^{2222}}{9^{1234}}\) \(\times\) \(\dfrac{1}{2^{988}}\)
A = \(\dfrac{7^{2222}}{9^{1234}\times2^{988}}\)
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