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1-2-3+4+5-6-7+8+...+197-198-199+200
=(1-2-3+4)+(5-6-7+8)+...+(197-198-199+200)
=0+0+...+0
=0
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\)
=\(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\)
=3-3\4-4x\(\frac{5}{2}\)
=\(\frac{5}{2}\)
a) 234 x 95 + 234 x 4 + 234 = 234 x ( 95 + 4 + 1 ) = 234 x 100 = 23400
a).234 x 95 + 234 x 4 + 234
= 234 x ( 95 + 4 + 1 )
= 234 x 100
= 23400
b) ( 1+1/2) x ( 1+1/3) x ( 1+1/4 )
= ( 2/2+1/2 ) x ( 3/3+1/3 ) x ( 4/4 + 1/4)
=3/2 + 4/3 + 5/4
= 18/12 + 16/12 + 20/12
= 54/12
= 9 /2
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
a/\(1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
b/\(1\dfrac{1}{2}:1\dfrac{1}{3}:1\dfrac{1}{4}\)
\(=\dfrac{3}{2}:\dfrac{4}{3}:\dfrac{5}{4}\)
\(=\dfrac{3}{2}\times\dfrac{3}{4}\times\dfrac{4}{5}\)
\(=\dfrac{9}{10}\)
#kễnh
a) \(=\dfrac{1\cdot2+1}{2}\cdot\dfrac{1\cdot3+1}{3}\cdot\dfrac{1\cdot4+1}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
b) \(=\dfrac{3}{2}:\dfrac{4}{3}:\dfrac{5}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{9}{10}\)
\(2\frac{3}{4}+1\frac{5}{6}+3\frac{1}{4}+2\frac{1}{6}\)
= \(\frac{11}{4}+\frac{11}{6}+\frac{13}{4}+\frac{13}{6}\)
= \(\left(\frac{11}{4}+\frac{13}{4}\right)+\left(\frac{11}{6}+\frac{13}{6}\right)\)
= \(\frac{24}{4}+\frac{24}{6}\)
= \(6+4\)
= \(10\)
1/2 + 1/3 + 1/4 + 1/5 + 4/5 + 3/4 + 2/3 + 1/2
= (1/2 + 1/2) + (1/3 + 2/3) + (1/4 + 3/4) + (1/5 + 4/5)
= 1 + 1 + 1 + 1
= 4
Ta đã biết \(\dfrac{1}{a\cdot a}< \dfrac{1}{\left(a+1\right)\left(a-1\right)}\) ( a ϵ Z )
⇒ \(Q=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{200\cdot200}\) < \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
Ta có \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
= \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
= \(\dfrac{1}{2}\left(1-\dfrac{1}{201}\right)=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}< \dfrac{100}{200}=\dfrac{1}{2}< \dfrac{3}{4}\)
Vậy Q < \(\dfrac{3}{4}\)