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ta có:

\(A=\left|-2,75\right|-3\frac{1}{3}+\frac{1}{4}\)

\(A=2,75-\frac{10}{3}+\frac{1}{4}\)

\(A=\frac{11}{4}+\frac{1}{4}-\frac{10}{3}\)

\(A=3-\frac{10}{3}\)

\(A=\frac{9}{3}-\frac{10}{3}=-\frac{1}{3}\)

\(B=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)\)

\(B=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}\)

\(B=-\frac{3}{5}-\frac{2}{5}=-1\)

MẶT khác: B=kA =>\(k=\frac{B}{A}\)

\(\Rightarrow k=\frac{-1}{-\frac{1}{3}}=1.3=3\)

vậy k=3

19 tháng 9 2015

2             

\(A=\dfrac{11}{4}-3-\dfrac{1}{2}+\dfrac{1}{4}=-\dfrac{1}{2}\)

\(B=\dfrac{-3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}=-1\)

Do đó: A>B

a: \(=\dfrac{2}{3}+\dfrac{38}{7}\cdot\dfrac{1}{2}-\dfrac{9}{7}+\dfrac{1}{12}\cdot16\)

\(=\dfrac{2}{3}+\dfrac{4}{3}+\dfrac{19}{7}-\dfrac{9}{7}=2+\dfrac{10}{7}=\dfrac{24}{7}\)

b: \(=\dfrac{11}{4}\cdot\dfrac{-2}{5}-\dfrac{11}{4}\cdot\dfrac{8}{5}+\dfrac{11}{4}\cdot\dfrac{-6}{5}\)

\(=\dfrac{11}{4}\cdot\dfrac{-16}{5}=\dfrac{-44}{5}\)

12 tháng 7 2018

2, 100^2+200^2+300^2+..+1000^2

=100^2+2^2×100^2+3^2×100^2+...+100^2×10^2

=100^2×( 1^2+2^2+3^2+..+10^2)

=100^2×385

= 3850000

20 tháng 7 2023

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

23 tháng 9 2020

a) \(\left(-\frac{3}{4}\right)^2:\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)

\(=\left[\left(-\frac{3}{4}\right):\frac{5}{4}\right]^2+\frac{147}{10}-\frac{34}{25}\)

\(=\left[\left(-\frac{3}{4}\right)\cdot\frac{4}{5}\right]^2+\frac{147}{10}-\frac{34}{25}\)

\(=\left(-\frac{3}{5}\right)^2+\frac{147}{10}-\frac{34}{25}=\frac{9}{25}+\frac{147}{10}-\frac{34}{25}=\left(\frac{9}{25}-\frac{34}{25}\right)+\frac{147}{10}=-1+\frac{147}{10}=\frac{137}{10}\)

b) \(\left(2\frac{1}{3}-1,5\right):\left(-6\frac{1}{6}+5\frac{1}{2}\right)+2,75\)

\(=\left(\frac{7}{3}-\frac{3}{2}\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}\)

\(=\frac{5}{6}:\left(-\frac{2}{3}\right)+\frac{11}{4}=\frac{5}{6}\cdot\left(-\frac{3}{2}\right)+\frac{11}{4}=-\frac{5}{4}+\frac{11}{4}=\frac{3}{2}\)

23 tháng 9 2020

                                                                Bài giải

\(a,\text{ }\left(-\frac{3}{4}\right)^2\text{ : }\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)

\(=\frac{9}{16}\text{ : }\frac{25}{16}+\frac{147}{10}-\frac{34}{25}\)

\(=\frac{18}{50}+\frac{735}{50}-\frac{68}{50}\)

\(=\frac{685}{50}=\frac{137}{10}\)