\(\frac{5932+6001x5931}{5932x6001-69}=\frac{6001x5931+5932}{5931x6001+6001-69}=\frac{6001x5931+5932}{6001x5931+5932}=1\)

đây nhé bạn 

nhân tiện cho mình cả lời giải lun nhé

\(=\dfrac{5931+6001\cdot5931+1}{5931\cdot6001+6001-69}\)

\(=\dfrac{5931\cdot6001+5932}{5931\cdot6001+5932}=1\)

C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)

C= CÂU HỎI TƯƠNG TỰ 

=> đcpm 

\(A=\frac{254\cdot399-145}{254+399\cdot253}\)

\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)

\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)

\(A=\frac{253\cdot399+254}{254+399\cdot253}\)

\(A=1\)

\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)

\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)

\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)

\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)

\(B=1\)

\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)

\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)

\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)

\(C=\frac{1}{2017}\)

câu 1: 1997,1997+1998,1988+1999,1999=5994,5994

câu 2: 5932+6001x5931/5932x6001-69=360117932

tck mình nha

1, 23/27 > 22/29

   12/25 < 25/49

\(\dfrac{5932+6001x5931}{5932x600-69}=\dfrac{5932+6001x\left(5932-1\right)}{5932x6001-69}=\dfrac{5932+6001x5932-6001}{5932x6001-69}=\dfrac{5932x6001-\left(6001-5932\right)}{5932x6001-69}=\dfrac{5932x6001-69}{5932x6001-69}=1\)

Tính nhanh:

a) \(\frac{254.399-145}{254+399.253}\)

\(=\)\(\frac{\left(253+1\right).399-145}{254+399.253}\)

\(=\)\(\frac{253.399+399-145}{254+399.253}\)

\(=\)\(\frac{253.399+254}{254+399.253}\)

\(=\)\(1\)

b) \(\frac{5932+6001.5931}{5932.6001-69}\)

\(=\)\(\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)

\(=\)\(\frac{5932+6001.5931}{5931.6001+6001-69}\)

\(=\)\(\frac{5932+6001.5931}{5932.6001+5932}\)

\(=\)\(1\)