Thu gọn biểu thức Q= 3^1 + 3^2 + 3^3 + … + 3^200
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NH
1
23 tháng 1 2017
\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\)
\(2B=2\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\right)\)
\(2B=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{200}{2^{199}}\)
\(2B-B=\left(2+\frac{3}{2^2}+...+\frac{200}{2^{199}}\right)-\left(1+\frac{3}{2^3}+...+\frac{200}{2^{200}}\right)\)
.... đặt A=... giiả tiếp
NH
0
1 tháng 10 2018
a) Ta có: A = 1 + 3 + 32 + ... + 3200
⇒ 3A = 3 + 32 + 33 + ... + 3200
⇒ 3A - A = (3 + 32 + 33 + ... + 3201) - (1 + 3 + 32 + ... + 3200)
⇒ 2A = 3201 - 1
⇒ A = 2A : A = \(\dfrac{3^{201}-1}{2}\)
Vậy A = \(\dfrac{3^{201}-1}{3}\)
ML
30 tháng 5 2016
Q=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\).\(\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{3-2}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\).\(\left(\sqrt{3}-\sqrt{2}\right)^{ }\)=-1
31 tháng 3 2022
B= -1/3+1/3^2-1/3^3+…+1/3^100-1/3^101
3B= -1+1/3-1/3^2+…+1/3^99-1/3^100
3B+B=4B=-1-1/3^101
=>B=(-1-1/3^101)/4
Vậy B=(-1-1/3^101)/4
20 tháng 8 2020
Ta có: \(E=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
\(=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)
\(=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\)
\(=-\frac{1}{\sqrt{3}-\sqrt{2}}\cdot\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\)
\(=-\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\cdot\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)
\(=-\sqrt{\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}}\)
\(=-\sqrt{\frac{1}{3-2}}=-1\)
NT
4
SL
22 tháng 5 2017
\(A=\sqrt{\left(3+\sqrt{3}\right)+2\sqrt{\left(3+\sqrt{3}\right)\left(\sqrt{5}-2\right)+\left(\sqrt{5}-2\right)}-\sqrt{3+\sqrt{3}}}\)
\(=\sqrt{3+\sqrt{3}}+\sqrt{\sqrt{5}-2}-\sqrt{3+\sqrt{3}}=\sqrt{\sqrt{5}-2}\)
ok???
\(Q=3^1+3^2+3^3+...+3^{200}\\3\cdot Q=3^2+3^3+3^4+...+3^{201}\\3Q-Q=(3^2+3^3+3^4+...+3^{201})-(3^1+3^2+3^3+...+3^{200})\\2Q=3^{201}-3\\\Rightarrow Q=\dfrac{3^{201}-3}{2}\)
\(Q=3+3^2+...+3^{200}\)
\(3Q=3\cdot\left(3+3^2+...+3^{200}\right)\)
\(3Q=3^2+3^3+...+3^{201}\)
\(3Q-Q=3^2+3^3+...+3^{201}-3-3^2+...-3^{200}\)
\(2Q=3^{201}-3\)
\(Q=\dfrac{3^{201}-3}{2}\)