Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\frac{5932+6001.5931}{5932.6001-69}\)
\(=\)\(\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\)\(\frac{5932+6001.5931}{5931.6001+6001-69}\)
\(=\)\(\frac{5932+6001.5931}{5932.6001+5932}\)
\(=\)\(1\)
\(a,\frac{154}{253}\)
\(b,\frac{5931}{69}\)
#)Giải :
A, \(\frac{254x399-145}{254+399x253}\)
\(=\frac{253x399+399-145}{254+399x253}\)
\(=\frac{253x399+254}{254+399x253}\)
\(=1\)
B, \(\frac{5932+6001x5931}{5931x6001-69}\)
\(=\frac{5932+6001x5931}{\left(5931+1\right)x6001-69}\)
\(=\frac{5932+6001x5931}{5931x6001+6001-69}\)
\(=\frac{5932+6001x5932}{5932x6001+5932}\)
\(=1\)
#~Will~be~Pens~#
\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)
\(=1\)
\(\frac{5932+6001x5931}{5932x6001-69}=\frac{6001x5931+5932}{5931x6001+6001-69}=\frac{6001x5931+5932}{6001x5931+5932}=1\)
C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)
C= CÂU HỎI TƯƠNG TỰ
=> đcpm
\(A=\frac{254\cdot399-145}{254+399\cdot253}\)
\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)
\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)
\(A=\frac{253\cdot399+254}{254+399\cdot253}\)
\(A=1\)
\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)
\(B=1\)
\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)
\(C=\frac{1}{2017}\)