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Chắc có lẽ bạn định làm như này:
\(\frac{1}{\left(3n+2\right)\left(3n+5\right)}=\frac{3}{3\left(3n+2\right)\left(3n+5\right)}=\frac{\left(3n+5\right)-\left(3n+2\right)}{3\left(3n+2\right)\left(3n+5\right)}=\frac{1}{3}\left[\frac{1}{3n+2}-\frac{1}{3n+5}\right]\)
\(A=\left(1+\frac{2}{4}\right)\left(1+\frac{2}{10}\right)...\left(1+\frac{2}{n^2+3n}\right)\)
\(A=\left(\frac{6}{4}\right)\left(\frac{12}{10}\right)...\left(\frac{n^2+3n+2}{n^2+3n}\right)\)
\(A=\left(\frac{2.3}{1.4}\right)\left(\frac{3.4}{2.5}\right)\left(\frac{4.5}{3.6}\right)...\left(\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+3\right)}\right)\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{3.4.5...\left(n+2\right)}{4.5.6...\left(n+3\right)}=\left(n+1\right).\frac{3}{\left(n+3\right)}=\frac{3\left(n+1\right)}{n+3}\)
Do \(0< n+1< n+3\Rightarrow\frac{n+1}{n+3}< 1\Rightarrow\frac{3\left(n+1\right)}{n+3}< 3\)
Vậy \(A< 3\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
ta có 3A = 3/1.4 + 3/4.7 + ... + 3/(3n-2).(3n+1)
3A = 1-1/4 + 1/4 - 1/7 +....+ 1/(3n-2) - 1/(3n+1)
3A = 1- 1/(3n+1)
Mà 1/(3n+1) > 0 suy ra 3A < 1 suy ra A<1/3
tk giúp mình nha
Đặt A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)
Ta có : A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)
= \(\frac{6}{4}.\frac{12}{10}.\frac{20}{18}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)
= \(\frac{3.2}{4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)
= \(\frac{3.2.3.4.4.5....n}{2.3.4.5.6.....\left(n+2\right)}\)
= \(\frac{3.\left(n+1\right)}{n+2}\)
Vậy A = \(\frac{3.\left(n+1\right)}{n+2}\)
a,
\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\\ =\left(n^2+3n-1\right)n+\left(n^2+3n-1\right)2-n^3+2\\ =n^3+3n^2-n+2n^2+6n-2-n^3+2\\ =5n^2+5n\\ =5\cdot\left(n^2+n\right)⋮5\\ \RightarrowĐpcm\)
b,
\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\\ =\left(6n+1\right)n+\left(6n+1\right)5-\left(3n+5\right)2n-\left(3n+5\right)\\ =6n^2+n+30n+5-6n^2-10n-3n-5\\ =18n⋮2\\ \RightarrowĐpcm\)
Quy đồng lên rồi tính bình thường thôi bạn