10+2*x=4mu 5/4
K
Khách
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PN
0
MS
2
4 tháng 9 2017
ban viet lai de bai duoc ko minh ko hieu neu viet so mu thi an shift roi an so 7 la mu vi du 3^2
TB
0
NN
2
6 tháng 5 2017
Ta thấy :
\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)
\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)
..............
\(\dfrac{1}{99^2}>\dfrac{1}{99.100}\)
\(\Rightarrow\) \(K>\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)
Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức ta có :
\(\dfrac{1}{4.5}=\dfrac{1}{4}-\dfrac{1}{5}\)
\(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
.......................
\(\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow\) \(K>\dfrac{1}{4}-\dfrac{1}{100}\)
\(\Rightarrow K>\dfrac{6}{25}>\dfrac{1}{5}\Rightarrow dpcm\) (1)
Ta có :
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
................
\(\dfrac{1}{99^2}< \dfrac{1}{98.99}\)
Dựa vào công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\) ta có :
\(K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{98.99}\)
\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.......+\dfrac{1}{98}-\dfrac{1}{99}\)
\(\Rightarrow\) \(K< \dfrac{1}{3}-\dfrac{1}{99}\)
Vậy \(K< \dfrac{32}{99}< \dfrac{1}{3}\Rightarrow dpcm\) (2)
(1) ; (2) \(\Rightarrow\) \(\dfrac{1}{5}< K< \dfrac{1}{3}\)
Ai thấy đúng thì ủng hộ nha !!!
6 tháng 5 2017
Công thức tổng quát: \(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
=>\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}< K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)
=>\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
=>\(\dfrac{1}{4}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{100}\)
=>\(\dfrac{1}{4}< K< \dfrac{1}{3}\)
=>\(\dfrac{1}{5}< K< \dfrac{1}{3}\left(do\dfrac{1}{4}>\dfrac{1}{5}\right)\)
Q
0
H9
HT.Phong (9A5)
CTVHS
2 tháng 11 2023
a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{7}\times1\)
\(=\dfrac{5}{7}\)
b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)
\(=1+0\)
\(=1\)
c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)
\(=\dfrac{5}{7}+\dfrac{2}{7}\)
\(=1\)
d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)
\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)
\(=\dfrac{1}{2}+1\)
\(=\dfrac{3}{2}\)
e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)
\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{12}{10}\)
\(=\dfrac{4}{5}+\dfrac{6}{5}\)
\(=\dfrac{10}{5}\)
\(=2\)
g) \(362\times728+326\times272\)
\(=326\times\left(728+272\right)\)
\(=326\times1000\)
\(=326000\)
CM
24 tháng 1 2017
Hướng dẫn giải:
a) 3 x (20 – 5)
Cách 1:
3 x (20 – 5) = 3 x 15 = 45
Cách 2:
3 x (20 – 5) = 3 x 20 – 3 x 5 = 60 – 15 = 45
b) 20 x (40 – 1)
Cách 1:
20 x (40 – 1) = 20 x 39 = 780
Cách 2:
20 x (40 – 1) = 20 x 40 – 20 x 1 = 800 – 20 = 780
\(10+2\times x=4^5:4\)
\(10+2\times x=256\)
\(2\times x=256-10\)
\(2\times x=246\)
\(x=246:2\)
\(x=123\)