A = \(\dfrac{254\times399-145}{254+399\times253}\)

A = \(\dfrac{\left(253+1\right)\times399-145}{254+399\times253}\)

A = \(\dfrac{253\times399+399-145}{254+399\times253}\)

A = \(\dfrac{253\times399+254}{254+399\times253}\)

A = 1 

a) \(254x399-145=\left(250+4\right)\left(400-1\right)-145\)

\(=100000+1600-250-145-4=101600-\left(250+145+4\right)\)

\(=101600-399=101201\)

b) \(254+399x253=254+\left(400-1\right)x\left(250+3\right)\)

\(=254+100000+1200-250-3\)

\(=101454-250-3=101201\)

\(\dfrac{a}{b}=1\)

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

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

\(=\frac{253.399+254}{254+399.253}=1\)

Trả lời

254.399-145/254+399.253

=253.399+399-145/254+399.253

=253.399+254/254+399.253

=1.

= 202292,42913385826771653543307087.............

=202292,4291338582677165353307087

\(\frac{254.399-145}{254+399.253}=\frac{\left(253+1\right).399-145}{254+399.253}=\frac{253.399+399-145}{254+399.253}=\frac{253.399+254}{254+399.253}=1\)

bằng 1 đó

= 1 nha

101201/101201

ghép đầu cuối dăt chung 399 ra ngoài 

kq vẫn như cũ chắc chắn 100% 

202292.4291

\(\frac{254+399.253}{399.254-145}\)

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

\(=\frac{254+399.253}{399.253+399.1-145}\)

\(=\frac{254+1}{1+399-145}=1\)

Chúc bạn học tốt!!!

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

C= CÂU HỎI TƯƠNG TỰ 

=> đcpm 

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

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

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

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

\(A=1\)

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

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

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

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

\(B=1\)

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

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

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

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