Ta có:
\(2n:\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+...+n}\right)=2020\)
<=> \(2n:\left(\frac{2}{2}+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{\left(n+1\right).n}\right)=2020\)
<=> \(n:\left(1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=2020\)
<=> \(n:\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=2020\)
<=> \(n:\left(1-\frac{1}{n+1}\right)=2020\)
<=> \(n:\frac{n}{n+1}=2020\)
<=> n + 1 = 2020
<=> n = 2019
Tìm n\(\in\)N* biết: \(2n\div\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+n}\right)=2020\)
Tim \(x\inℕ^∗\)\(2x:\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}\right)=2020\)
Chứng minh rằng với mọi n \(\inℕ^∗\):
D = \(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}< 1\)
F = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
Bài 1: CMR
a) A = \(\frac{\left(n+1\right).\left(n+2\right)....\left(2n-1\right).\left(2n\right)}{2^n}\) là số nguyên.
b) B = \(\frac{3.\left(n+1\right).\left(n +2\right)...\left(3n-1\right).3n}{3^n}\)là số nguyên.
bài 1 : tìm x: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x : \(\left(8x-1\right)^{2n+1}=5^{2n+1}\)(n thuộc N)
Tính :
1 + 3 + 5 + 7 + ... + (2n - 1) = 225
Giải :
Theo công thức tính dãy số , ta có :
\(\frac{\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}.\left[\left(2n-1\right)+1\right]}{2}=225\)
\(\frac{\left\{\left[2n-2\right]:2+1\right\}.2n}{2}=225\)
\(\left\{\left[2n-2\right]:2+1\right\}.n=450\)(Lượt giản thừa số 2)
\(\left\{\frac{2n-2}{2}+1\right\}.n=225\)
\(\left\{\frac{2n-2}{2}+\frac{2}{2}\right\}.n=225\)
\(\frac{2n-2+2}{2}.n=225\)
\(\frac{2n}{2}.n=225\)
\(n^2=225\)
\(\Rightarrow n=\sqrt{225}=15\)
CMR với mọi số tự nhiên n>2 thì :
a)\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)<\(\frac{1}{2}\)
b)\(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}\)<\(\frac{1}{4}\)
c)\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{\left(2n+1\right)^2}\right)\)<2
CÁC BÀI TOÁN VỀ LŨY THỪA SỐ HỮU TỈ
Tìm x biết:
a, \(\left(5x+1\right)^2=\frac{36}{49}\)
b, \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
c, \(\left(8x-1\right)^{2n+1}=5^{2n+1}\) (n thuộc N) CÁC BN NHỚ GIẢI THEO CÁCH CỦA LỚP 7 NHÉ!!! ^=^
bài 1 tìm x, y \(\inℕ^∗\) biết
\(\frac{1}{x}-\frac{1}{y}=\frac{1}{x-y}\)
bài 2 cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
CMR \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
bài 3
A=\(\frac{\left(1+2+3+.....+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.6,3\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{100}}\)
mình đang cần gấp mình chỉ cần 1 bài thôi cũng đc
