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

Câu b 

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

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

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

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

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

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

\(=\frac{6001x5932-69}{5932x6001-69}\)

\(=1\)

\(\frac{5932+6001\times5931}{5932\times6001-69}\)

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

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

\(=\frac{5932+6001\times5931}{5931\times6001+5932}\)

\(=1\)

~ Study Well ~

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

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

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

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

                                    \(=\frac{253.399+1.399-145}{254+399.253}\)

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

                                    \(=1.\)

\(\frac{5932+6001.5931}{5932.6001-69}=\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)

                                    \(=\frac{5932+6001.5931}{5931.6001+1.6001-69}\)

                                    \(=\frac{5932+6001.5931}{5931.6001+5932}\)

                                    \(=1.\)

Câu thứ nhất là: 202292,4291.       ;       Câu thứ hai là: 36011793,2.

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

a, \(\dfrac{254\times399-145}{254+399\times253}\)

= \(\dfrac{\left(253+1\right)\times399-`45}{254+399\times253}\)

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

= \(\dfrac{253\times399+254}{253\times399+254}\)

= 1

b, \(\dfrac{5932+6001\times5931}{5932\times6001-69}\)

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

= \(\dfrac{5932+6001\times5931}{5931\times6001+6001-69}\)

= \(\dfrac{5932+6001\times5931}{5931\times6001+5932}\)

= 1

a,109/508

b,1

A= 109/508

B=1

A) 0,996

B) 1

Bài của mình làm đúng 100% luôn nha bạn,cứ yên tấm đi không sai đâu

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

C= CÂU HỎI TƯƠNG TỰ 

=> đcpm 

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

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

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

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

\(A=1\)

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

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

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

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

\(B=1\)

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

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

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

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