Chứng minh: 3/2*5+ 3/5*8 + 3/8*11 + 3/11*14 + 3/14*17 <1/2
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\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(=\)\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk giúp !!
a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)
Vậy A<\(\frac{1}{2}\).
b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< 1-\frac{1}{100}< 1\)
Vậy \(B< 1\).
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
b: \(27D=3^{14}+3^{17}+...+3^{2024}\)
\(\Leftrightarrow26D=3^{2024}-3^{11}\)
hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)
c: \(25E=-5^4-5^6-...-5^{1002}\)
\(\Leftrightarrow24E=-5^{1002}+5^2\)
hay \(E=\dfrac{-5^{1002}+5^2}{24}\)
a: \(=3\cdot\dfrac{6+14-9}{42}\cdot\dfrac{14}{11}\)
\(=3\cdot\dfrac{11}{42}\cdot\dfrac{14}{11}=1\)
b: \(=\dfrac{19}{5}\cdot\dfrac{11}{10}+\dfrac{9}{5}\cdot\dfrac{7}{3}\)
\(=\dfrac{209}{50}+\dfrac{63}{15}=4.18+4.2=8,38\)
Lời giải chi tiết:
11 + 8 = 19 | 14 + 2 = 16 | 7 + 3 = 10 | 12 + 5 = 17 |
19 – 8 = 11 | 16 – 2 = 14 | 10 – 3 = 7 | 17 – 5 = 12 |
Phương pháp giải:
Tính nhẩm các số rồi điền kết quả vào chỗ trống.
Lời giải chi tiết:
14 − 9 = 5 | 16 − 7 = 9 | 12 − 8 = 4 | 6 + 9 = 15 |
8 + 8 = 16 | 11 − 5 = 6 | 13 − 6 = 7 | 18 − 9 = 9 |
14 − 6 = 8 | 17 − 8 = 9 | 12 − 5 = 7 | 3 + 9 = 12 |
9 + 4 = 13 | 11 − 9 = 2 | 3 + 8 = 11 | 16 − 8 = 8 |
Tính nhẩm:
14 − 9 = ..... | 16 − 7 = ..... | 12 − 8 = ..... | 6 + 9 = ..... |
8 + 8 = ..... | 11 − 5 = ..... | 13 − 6 = ..... | 18 − 9 = ..... |
14 − 6 = ..... | 17 − 8 = ..... | 12 − 5 = ..... | 3 + 9 = ..... |
9 + 4 = ..... | 11 − 9 = ..... | 3 + 8 = ..... | 16 − 8 = ..... |
a; \(\dfrac{9}{4}\) - \(\dfrac{-11}{4}\)
= \(\dfrac{9}{4}\) + \(\dfrac{11}{4}\)
= \(\dfrac{20}{4}\)
= 5
b; \(\dfrac{7}{8}\) - \(\dfrac{3}{-8}\) - \(\dfrac{1}{8}\)
= \(\dfrac{7}{8}\) + \(\dfrac{3}{8}\) - \(\dfrac{1}{8}\)
= \(\dfrac{7+3-1}{8}\)
= \(\dfrac{9}{8}\)
c; \(\dfrac{-5}{21}\) - \(\dfrac{25}{21}\) - \(\dfrac{-1}{21}\)
= \(\dfrac{-5}{21}\) - \(\dfrac{25}{21}\) + \(\dfrac{1}{21}\)
= \(\dfrac{-5-25+1}{21}\)
= \(\dfrac{-29}{21}\)
\(A=\frac{2}{7}+\frac{-3}{8}+\frac{11}{7}+\frac{1}{3}+\frac{1}{7}+\frac{5}{-8}\)
\(A=\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)+\left(\frac{-3}{8}+\frac{5}{-8}\right)+\frac{1}{3}\)
\(A=2-1+\frac{1}{3}\)
\(A=\frac{4}{3}\)
\(B=\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+17\)
\(B=\left(\frac{3}{17}+\frac{14}{17}\right)+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=1+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=18+\frac{-5}{13}+\frac{-18}{35}\)
\(B=\frac{7781}{455}\)