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Đặt 317=a; 111=b
Theo đề, ta có: \(2\dfrac{1}{a}\cdot\dfrac{3}{b}-\dfrac{a-1}{a}\cdot\dfrac{1}{b}-\dfrac{4}{ab}\)
\(=\dfrac{3\left(2a+1\right)}{ab}-\dfrac{a-1}{ab}-\dfrac{4}{ab}\)
\(=\dfrac{6a+3-a+1-4}{ab}=\dfrac{5a}{ab}=\dfrac{5}{b}=\dfrac{5}{111}\)
1) Đặt \(\frac{1}{317}=a;\frac{3}{111}=b\) thế vào mà làm thôi
mấy câu sau tương tự
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 a=4453, b=1997
Ta có: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+
\(=\dfrac{5a+6-4a-6}{ab}\)
\(=\dfrac{1}{b}\)
\(=\dfrac{1}{1997}\)
Đặt S= \(2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315.651}+\dfrac{4}{105}\)
= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+\dfrac{651-1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+1-\dfrac{1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
Đặt \(\dfrac{1}{315}=a,\dfrac{1}{651}=b\)
\(\Rightarrow S=\left(2+a\right).b-3a.\left(4-b\right)-4ab+12a\)
\(=2b+ab-12a+3ab-4ab+12a\)
\(=2b=\dfrac{2}{651}\)
Đặt \(\dfrac{1}{315}=x,\dfrac{1}{651}=y\)
\(\Rightarrow A=\left(2+x\right)y-3x\left(4-y\right)-4xy+12x\)
\(=2y+xy-12x+3xy-4xy+12x\)
\(=2y\)
Thay \(y=\dfrac{1}{651}\Rightarrow A=\dfrac{2}{651}\)
Vậy...
Đặt \(A=2\dfrac{1}{317}.\dfrac{3}{111}-\dfrac{316}{317}.\dfrac{1}{111}-\dfrac{4}{317.111}\)
\(=\left(2+\dfrac{1}{317}\right).\dfrac{3}{111}-\left(1-\dfrac{1}{317}\right).\dfrac{1}{111}-4.\dfrac{1}{317}.\dfrac{1}{111}\)
\(=6.\dfrac{1}{111}+3.\dfrac{1}{317}.\dfrac{1}{111}-\dfrac{1}{111}+\dfrac{1}{317}.\dfrac{1}{111}-4.\dfrac{1}{317}.\dfrac{1}{111}\)
Đặt \(a=\dfrac{1}{111};b=\dfrac{1}{317}\). Khi đó
\(A=6a+3ab-a+ab-4ab=5a=\dfrac{5}{111}\)
Vậy A=\(\dfrac{5}{111}\)