Câu hỏi của Lê Hoàng Tiến Đạt

Question

tính nhanh$\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)$

Nguyễn Văn Hưng A · Accepted Answer

$\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)$$=\frac{1999}{2011}-\frac{2011}{1999}-\frac{12}{1999}+\frac{12}{2011}$$=(\frac{1999}{2011}+\frac{12}{2011})-(\frac{2011}{1999}-\frac{12}{1999})$$=1-1=0$Câu trả lời: $\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)$$=\frac{1999}{2011}-\frac{2011}{1999}-\frac{12}{1999}+\frac{12}{2011}$$=(\frac{1999}{2011}+\frac{12}{2011})-(\frac{2011}{1999}-\frac{12}{1999})$$=1-1=0$

Gukmin · Answer

$\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)$$=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}$$=\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(\frac{-2011}{1999}+\frac{12}{1999}\right)$$=1-1$$=0$LinzCâu trả lời: $\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)$$=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}$$=\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(\frac{-2011}{1999}+\frac{12}{1999}\right)$$=1-1$$=0$Linz

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