\(\dfrac{4}{15}\):\(\dfrac{4}{7}\)<x<\(\dfrac{2}{5}\)x\(\dfrac{10}{3}\)
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a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
`@` `\text {Ans}`
`\downarrow`
`j)`
`-3/4 + 2/7 + (-1)/4 + 3/5 + 5/7`
`= (-3/4 - 1/4) + (2/7 + 5/7) + 3/5`
`= -1 + 1 + 3/5`
`= 3/5`
`k)`
`-2/17 + 15/23 + (-15)/17 + 4/19 + 8/23`
`= (-2/17 - 15/17) + (15/23 + 8/23) + 4/19`
`= -1 + 1 + 4/19`
`= 4/19`
$#KDN040510$
j: =-3/4-1/4+2/7+5/7+3/5
=-1+1+3/5
=3/5
k: =-2/17-15/17+15/23+8/23+4/19
=-1+1+4/19
=4/19
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=2.\dfrac{3}{16}\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
1: \(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}\right)+\dfrac{16}{15}\left(\dfrac{4}{7}-\dfrac{5}{9}\right)\)
\(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}+\dfrac{4}{7}-\dfrac{5}{9}\right)=0\)
2: \(=\dfrac{29}{9}\left(15+\dfrac{4}{7}-8-\dfrac{1}{7}+\dfrac{15}{7}-\dfrac{1}{7}\right)\)
\(=\dfrac{20}{9}\cdot\left(7\cdot\dfrac{18}{7}\right)=\dfrac{20}{9}\cdot18=40\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
a) \(\dfrac{-4}{6}\)
b) \(\dfrac{1}{3}\)-\(\dfrac{20}{60}\)
= 0
c) \(-20^2\)+(\(-50^3\))
= -400 + (-125000)
=-125400
d) \(\dfrac{-2}{7}\)+\(\dfrac{15}{35}\)
= \(\dfrac{-10}{35}\)+\(\dfrac{15}{35}\)
= \(\dfrac{5}{35}\)=\(\dfrac{1}{7}\)
A = \(\dfrac{4}{3}\) . \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) . \(\dfrac{4}{11}\) + \(\dfrac{4}{11}\) . \(\dfrac{4}{15}\) + ... + \(\dfrac{4}{95}\) . \(\dfrac{4}{99}\)
A = \(\dfrac{4.4}{3.7}\) + \(\dfrac{4.4}{7.11}\) + \(\dfrac{4.4}{11.15}\) + ... + \(\dfrac{4.4}{95.99}\)
A = \(\dfrac{16}{3.7}\) + \(\dfrac{16}{7.11}\) + \(\dfrac{16}{11.15}\) + ... + \(\dfrac{16}{95.99}\)
A = 4.( \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) + ... + \(\dfrac{4}{95.99}\))
A = 4.( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\))
A = 4.(\(\dfrac{1}{3}\) - \(\dfrac{1}{99}\))
A = 4.(\(\dfrac{33}{99}\) + \(\dfrac{-1}{99}\))
A = 4. \(\dfrac{32}{99}\)
A = \(\dfrac{4.32}{99}\)
A = \(\dfrac{128}{99}\)
Vậy A = \(\dfrac{128}{99}\)