a) 254x399-145 254+399x253 b) 5932+6001x5931 5932x6001x69
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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) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)
\(\frac{254.399-145}{254+399.253}=\frac{\left(253+1\right).399-145}{254+399.253}\)
\(=\frac{253.399+1.399-145}{254+399.253}\)
\(=\frac{253.399+254}{254+399.253}\)
\(=1.\)
\(\frac{5932+6001.5931}{5932.6001-69}=\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\frac{5932+6001.5931}{5931.6001+1.6001-69}\)
\(=\frac{5932+6001.5931}{5931.6001+5932}\)
\(=1.\)
Câu 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
= \(\frac{35597863}{35597833}=1,000000843\)
\(\frac{5932+6001x5931}{5932x6001-69}=\frac{6001x5931+5932}{5931x6001+6001-69}=\frac{6001x5931+5932}{6001x5931+5932}=1\)
254 x 399 - 145/254 + 339 x 253
= 101346 - 145/254 + 85767
= 187112,4291
\(\frac{254\cdot399-145}{254+399\cdot253}=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)
\(=\frac{253\cdot399+399-145}{254+399\cdot253}\)
\(=\frac{253\cdot399+254}{254+399\cdot253}\)
\(=1\)
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) \(254+399x253=254+\left(400-1\right)x\left(250+3\right)\)
\(=254+100000+1200-250-3\)
\(=101454-250-3=101201\)
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