Question

[Trần Thảo Nguyên]

Bài 6: Tính nhanh: 

a) \(\dfrac{254x399-145}{254+399x253}\)                      ( x ở trong bài này là nhân ạ)

b) \(\dfrac{5932+6001x5931}{5932x6001-69}\)

c) \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)  + \(\dfrac{1}{8}\)  + \(\dfrac{1}{16}\)  + \(\dfrac{1}{32}\)

Answer 1

[Estelle SEM]

uh là nhân

Answer 2

[Nguyễn Đức Trí]

a) \(=\dfrac{254x\left(400-1\right)-145}{254+\left(400-1\right)x253}=\dfrac{254x400-254-145}{254+253x400-253}\)

\(=\dfrac{101600-399}{101200+1}=\dfrac{101211}{101201}=\dfrac{101201+10}{101201}=1+\dfrac{10}{101201}\)

b) \(=\dfrac{5392+\left(600+1\right)x5391}{5392x\left(600+1\right)-69}=\dfrac{5392+600x5391+5391}{5392x600+5392-69}\)

\(=\dfrac{10783+3234600}{3235200+5323}=\dfrac{\text{3245383}}{\text{3240523}}=\dfrac{3240523+60}{3240523}=1+\dfrac{60}{3240523}\)

c) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)