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a)
\(=\frac{3}{5}.\frac{3}{7}+\frac{3}{5}.\frac{4}{7}-\left(1+\frac{3}{5}\right)\)
\(=\frac{3}{5}\left(\frac{3}{7}+\frac{4}{7}\right)-1-\frac{3}{5}\)
\(=\frac{3}{5}-1-\frac{3}{5}\)
\(=-1\)
b) \(=\frac{2^2.5.7.5^2.7^3}{2^2.5^2.7^{2.2}}\)
\(=\frac{2^2.5^{1+2}.7^{3+1}}{2^2.5^2.7^4}=\frac{2^2.5^3.7^4}{2^2.5^2.7^4}=2^{2-2}.5^{3-2}.7^{4-4}=2^0.5^1.7^0=1.5.1=5\)
Lời giải:
$x(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7})< 1\frac{6}{7}$
$x(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7})< \frac{13}{7}$
$x(1-\frac{1}{7})< \frac{13}{7}$
$x.\frac{6}{7}< \frac{13}{7}$
$x< \frac{13}{7}: \frac{6}{7}=\frac{13}{6}$
Vì $x$ là số nguyên nên $x\leq 2$
Vậy $x$ là các số nguyên sao cho $x\leq 2$.
\(A=\left[0,8\cdot7+(0,8)^2\right]\cdot\left[1,25\cdot7-\frac{4}{5}\cdot1,25\right]-47,86\)
\(=0,8\cdot(7+0,8)\cdot1,25\cdot(7-0,8)-47,86\)
\(=0,8\cdot7,8\cdot1,25\cdot6,2-47,86\)
\(=48,36-47,86=0,5\)
\(B=\frac{(1,09-0,29)\cdot\frac{5}{4}}{(18,9-16,65)\cdot\frac{8}{9}}=\frac{0,8\cdot1,25}{2,25\cdot\frac{8}{9}}=\frac{1}{2}\)
\(A:B=0,5:\frac{1}{2}=\frac{1}{2}:\frac{1}{2}=\frac{1}{2}\cdot2=1\)
A gấp 1 lần B
\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(E=\frac{1}{1}-\frac{1}{99}\)
\(E=\frac{98}{99}\)
E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt
Đặt \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{n\left(n+2\right)}\)
\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{n\left(n+2\right)}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(2A=\frac{1}{3}-\frac{1}{n+2}\)
\(2A=\frac{n-1}{3\left(n+2\right)}\)
\(A=\frac{n-1}{6\left(n+2\right)}\)
Ta có : \(\frac{1}{2}=\frac{3\left(n+2\right)}{2\cdot3\left(n+2\right)}=\frac{3n+6}{6\left(n+2\right)}\)
Dễ thấy \(n-1< 3n+6\)
Do đó \(\frac{1}{2}>A\)
1/2×(1/3-1/5+1/5-1/7+.....+1/n-1/n+2)
=> 1/2×(1/3-1/n+2) <1/2
=> 1/3-1/n+2< 1
Vậy 1/3×5+1/5×7+....+1/n×n+2 < 1/2