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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) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)
C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)
C= CÂU HỎI TƯƠNG TỰ
=> đcpm
\(A=\frac{254\cdot399-145}{254+399\cdot253}\)
\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)
\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)
\(A=\frac{253\cdot399+254}{254+399\cdot253}\)
\(A=1\)
\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)
\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)
\(B=1\)
\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)
\(C=\frac{1}{2017}\)
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) \(\frac{5932+6001.5931}{5932.6001-69}\)
\(=\)\(\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\)\(\frac{5932+6001.5931}{5931.6001+6001-69}\)
\(=\)\(\frac{5932+6001.5931}{5932.6001+5932}\)
\(=\)\(1\)
Bài 2
a) \(x\times1\frac{1}{4}=3\frac{3}{4}\)
\(x\times\frac{5}{4}=\frac{15}{4}\)
\(x=\frac{15}{4}.\frac{4}{5}\)
\(x=3\)
b) \(x-\frac{3}{4}=6\times\frac{3}{8}\)
\(x-\frac{3}{4}=\frac{9}{4}\)
\(x=\frac{9}{4}+\frac{3}{4}\)
\(x=3\)
Những câu còn lại tương tự
#)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~#
\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)
\(=1\)
Câu a sai đề (sửa đề lại rồi làm tương tự câu b
Câu b
\(\frac{5932+6001\times5931}{5932\times6001-69}=\frac{5932+6001\times5931}{\left(5931+1\right)\times6001-69}=\frac{5932+6001\times5931}{5931\times6001+6001-69}=\frac{5932+6001\times5931}{5932\times6001+5932}=1\)
\(254.399-\frac{145}{254+399.255}=101346-\frac{145}{101999}=1012031170\)
\(5932+6001.\frac{5931}{35597863}=1\)
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
= \(\frac{35597863}{35597833}=1,000000843\)
\(\frac{5932+6001x5931}{5932x6001-69}=\frac{6001x5931+5932}{5931x6001+6001-69}=\frac{6001x5931+5932}{6001x5931+5932}=1\)