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câu 1: 1997,1997+1998,1988+1999,1999=5994,5994
câu 2: 5932+6001x5931/5932x6001-69=360117932
tck mình nha
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) \(\frac{5932+6001.5931}{5932.6001-69}\)
\(=\)\(\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\)\(\frac{5932+6001.5931}{5931.6001+6001-69}\)
\(=\)\(\frac{5932+6001.5931}{5932.6001+5932}\)
\(=\)\(1\)
58 x 42 + 32 x 8 + 5 x 16
= 2436 + 256 + 80
= 2692 + 80
= 2772
456 : 2 x 18 + 456 : 3 - 102
= 228 x 18 + 152 - 102
= 4104 + 152 - 102
= 4256 - 102
= 4154
( 254 x 399 - 145 ) : ( 254 + 399 x 253 )
= ( 101346 - 145 ) : ( 254 + 100947 )
= 101201 : 101201
= 1
( 5932 + 6001 x 5931 ) : ( 5932 x 6001 - 99 )
= ( 5932 + 35591931 ) : ( 35597932 - 99 )
= 35597863 : 35597833
= \(\frac{35597863}{35597833}=1,000000843\)
C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)
C= CÂU HỎI TƯƠNG TỰ
=> đcpm
\(A=\frac{254\cdot399-145}{254+399\cdot253}\)
\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)
\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)
\(A=\frac{253\cdot399+254}{254+399\cdot253}\)
\(A=1\)
\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)
\(B=1\)
\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)
\(C=\frac{1}{2017}\)
\(a,\frac{154}{253}\)
\(b,\frac{5931}{69}\)
\(A=\frac{5932+6001.5931}{5932.6001-69}=\frac{5932+6001.5931}{5931.6001+6001-69}=\frac{5932+6001.5931}{5931.6001+5932}=1\)