uh là nhân

a) \(=\dfrac{254x\left(400-1\right)-145}{254+\left(400-1\right)x253}=\dfrac{254x400-254-145}{254+253x400-253}\)

\(=\dfrac{101600-399}{101200+1}=\dfrac{101211}{101201}=\dfrac{101201+10}{101201}=1+\dfrac{10}{101201}\)

b) \(=\dfrac{5392+\left(600+1\right)x5391}{5392x\left(600+1\right)-69}=\dfrac{5392+600x5391+5391}{5392x600+5392-69}\)

\(=\dfrac{10783+3234600}{3235200+5323}=\dfrac{\text{3245383}}{\text{3240523}}=\dfrac{3240523+60}{3240523}=1+\dfrac{60}{3240523}\)

c) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)

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

Câu a sai đề (sửa đề lại rồi làm tương tự câu b

Câu b 

\(\frac{5932+6001\times5931}{5932\times6001-69}=\frac{5932+6001\times5931}{\left(5931+1\right)\times6001-69}=\frac{5932+6001\times5931}{5931\times6001+6001-69}=\frac{5932+6001\times5931}{5932\times6001+5932}=1\)

\(254.399-\frac{145}{254+399.255}=101346-\frac{145}{101999}=1012031170\)

\(5932+6001.\frac{5931}{35597863}=1\)

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

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

\(=1\)

ban ơi có đúng ko

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

\(=\frac{254+399\times253}{253\times399+254}=1\)

#)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~#

\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)

\(=1\)

\(\frac{254x399-145}{254+399x253}\)

=\(\frac{\left(253+1\right)x399-145}{254+399x253}\)

=\(\frac{253x399+399-145}{254+399x253}\)

=\(\frac{253.399+254}{254+399x253}\)=\(1\)

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

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

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

=\(\frac{5932+6001x5931}{5931x6001x5932}\)=1

# Hok tốt

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

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

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

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

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

bằng 1 đó

= 1 nha

101201/101201

ghép đầu cuối dăt chung 399 ra ngoài 

kq vẫn như cũ chắc chắn 100% 

202292.4291

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

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

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

Trả lời

254.399-145/254+399.253

=253.399+399-145/254+399.253

=253.399+254/254+399.253

=1.