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ta có:
\(A=\left|-2,75\right|-3\frac{1}{3}+\frac{1}{4}\)
\(A=2,75-\frac{10}{3}+\frac{1}{4}\)
\(A=\frac{11}{4}+\frac{1}{4}-\frac{10}{3}\)
\(A=3-\frac{10}{3}\)
\(A=\frac{9}{3}-\frac{10}{3}=-\frac{1}{3}\)
VÀ
\(B=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)\)
\(B=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}\)
\(B=-\frac{3}{5}-\frac{2}{5}=-1\)
MẶT khác: B=kA =>\(k=\frac{B}{A}\)
\(\Rightarrow k=\frac{-1}{-\frac{1}{3}}=1.3=3\)
vậy k=3
\(A=\dfrac{11}{4}-3-\dfrac{1}{2}+\dfrac{1}{4}=-\dfrac{1}{2}\)
\(B=\dfrac{-3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}=-1\)
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: \(=\dfrac{11}{4}\cdot\dfrac{-2}{5}-\dfrac{11}{4}\cdot\dfrac{8}{5}+\dfrac{11}{4}\cdot\dfrac{-6}{5}\)
\(=\dfrac{11}{4}\cdot\dfrac{-16}{5}=\dfrac{-44}{5}\)
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
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)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
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) \(\left(2\frac{1}{3}-1,5\right):\left(-6\frac{1}{6}+5\frac{1}{2}\right)+2,75\)
\(=\left(\frac{7}{3}-\frac{3}{2}\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}\)
\(=\frac{5}{6}:\left(-\frac{2}{3}\right)+\frac{11}{4}=\frac{5}{6}\cdot\left(-\frac{3}{2}\right)+\frac{11}{4}=-\frac{5}{4}+\frac{11}{4}=\frac{3}{2}\)