Rút gọn biểu thức sau:
\(A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+n}\right)\)với n lá các số tự nhiên lớn hơn 2.
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26 tháng 8 2017
Với mọi k thuộc N và k > 2 thì ta có :
\(1-\frac{1}{1+2+....+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=1-\frac{2}{k\left(k+1\right)}=\frac{k^2+k-2}{k\left(k+1\right)}=\frac{\left(k+2\right)\left(k-1\right)}{k\left(k+1\right)}\)
Áp dụng vào A ta được :
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)
\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)
4 tháng 10 2021
\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)
MH
4 tháng 10 2021
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2022
\(S\left(x\right)=\dfrac{1}{x^2}+\dfrac{2}{x^3}+...+\dfrac{n}{x^{n+1}}\)
\(\Rightarrow x.S\left(x\right)=\dfrac{1}{x}+\dfrac{2}{x^2}+\dfrac{3}{x^3}+...+\dfrac{n}{x^n}\)
\(\Rightarrow x.S\left(x\right)-S\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+...+\dfrac{1}{x^n}-\dfrac{n}{x^{n+1}}\)
\(\Rightarrow\left(x-1\right)S\left(x\right)=\dfrac{1}{x}.\dfrac{1-\left(\dfrac{1}{x}\right)^n}{1-\dfrac{1}{x}}-\dfrac{n}{x^{n+1}}=\dfrac{x^n-1}{x^n\left(x-1\right)}-\dfrac{n}{x^{n+1}}=\dfrac{x^{n+1}-x-n\left(x-1\right)}{x^{n+1}\left(x-1\right)}\)
\(\Rightarrow S\left(x\right)=\dfrac{x^{n+1}-\left(n+1\right)x+n}{x^{n+1}\left(x-1\right)^2}\)
2 tháng 7 2021
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
24 tháng 10 2023
a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)
\(=1-x\sqrt{x}-x\sqrt{x}\)
\(=1-2x\sqrt{x}\)
b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\left(\dfrac{\left(1-\sqrt{a}\right)\cdot\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{1}{\sqrt{a}+1}\right)^2\cdot\left(a+\sqrt{a}+1+\sqrt{a}\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
5 tháng 2 2022
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
5 tháng 2 2022
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)