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Đặt 3179=a; 1111=b
\(K=2\dfrac{1}{a}\cdot\dfrac{3}{b}-\dfrac{a-1}{a}\cdot\dfrac{1}{b}-\dfrac{4}{a\cdot b}\)
\(=\dfrac{2a+1}{a}\cdot\dfrac{3}{b}-\dfrac{a-1}{ab}-\dfrac{4}{ab}\)
\(=\dfrac{6a+3-a+1-4}{ab}\)
\(=\dfrac{4a}{ab}=\dfrac{4}{b}=\dfrac{4}{1111}\)
Đặt 3154=a; 6517=b
Theo đề, ta có:
\(B=2\dfrac{1}{a}\cdot\dfrac{3}{b}-\dfrac{3}{a}\cdot\dfrac{b-1}{b}+\dfrac{6}{\dfrac{1}{2}a}-\dfrac{6}{a\cdot b}\)
\(=\dfrac{2a+1}{a}\cdot\dfrac{3}{b}-\dfrac{3b-3}{ab}+\dfrac{12}{a}-\dfrac{6}{a\cdot b}\)
\(=\dfrac{6a+3-3b+3-6}{ab}+\dfrac{12b}{ab}\)
\(=\dfrac{6a-3b+12b}{ab}=\dfrac{6a+9b}{ab}=\dfrac{3\left(2a+3b\right)}{ab}\)
\(=\dfrac{3\left(2\cdot3154+3\cdot6517\right)}{3154\cdot6517}\)
1: \(A=4\dfrac{7}{1000}\cdot\dfrac{1}{999}-1\dfrac{1}{500}\cdot\dfrac{4}{999}+\dfrac{1001}{999\cdot1000}\)
Đặt 1/1000=a; 1/999=b
\(A=\left(4+7a\right)\cdot b-\left(1+2a\right)\cdot4b+b\cdot\dfrac{1001}{1000}\)
\(=4b+7ab-4b-8ab+b\cdot\left(1+a\right)\)
=-ab+b+b+ba=2b=2/999
2: Đặt 1/4587=a;1/3897=b
\(B=a\cdot\left(7+b\right)-\left(3+1-a\right)\cdot2b-7a-3ab\)
=7a+ab-8a+2ab-7a-3ab
=-8a=-8/4587
a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)
\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)
* là dấu nhân à bn ????
uk