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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=\(\dfrac{5}{111}\)
1) Đặt \(\frac{1}{317}=a;\frac{3}{111}=b\) thế vào mà làm thôi
mấy câu sau tương tự
a.\(\frac{5x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(1-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\) = \(\frac{5x+1-1+3x-2x^2+2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}\) =\(\frac{10x+2}{x^3-1}\)
b.\(\frac{5}{x+1}+\frac{10}{x^2-x+1}-\frac{15}{x^3+1}\)( đến đây dễ r đúng ko)
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
bạn ơi tại sao lại bằng 2x +6, bạn có thể giải đáp cho mình đc ko
a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)
Đặt 317=a; 111=b
Theo đề, ta có: \(2\dfrac{1}{a}\cdot\dfrac{3}{b}-\dfrac{a-1}{a}\cdot\dfrac{1}{b}-\dfrac{4}{ab}\)
\(=\dfrac{3\left(2a+1\right)}{ab}-\dfrac{a-1}{ab}-\dfrac{4}{ab}\)
\(=\dfrac{6a+3-a+1-4}{ab}=\dfrac{5a}{ab}=\dfrac{5}{b}=\dfrac{5}{111}\)