Câu hỏi của Vinh Manh

Question

cho biểu thức:A=[6$\times\left(-\frac{1}{3}\right)^2-3\times\left(-\frac{1}{3}\right)+1$]$\div\left(-\frac{1}{3}-1\right)$

B=\(\left(729-1^3\right)\times\left(729-3^3\right)\times...\times\left(7...

Nguyễn Huy Tú · Accepted Answer

Ta có: $A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)$

$=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)$

$=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)$

$=\frac{8}{3}.\left(-\frac{3}{4}\right)$

$=-2$

$B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)$

$\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)$

$\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)$

$\Rightarrow B=0$

Vì -2 < 0 nên A < B

Vậy A < BCâu trả lời: Ta có: $A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)$