tính nhanh
a.\(\frac{254x399-145}{254+399x253}\)
b. \(\frac{5932x6001x5931}{5932x6001-69}\)
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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) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)
a . 254 x 399 - 145 / 254 + 399 x 253 = ( 253 + 1 ) x 399 - 145 / 254 + 399 x 253
= 253 x 399 + 1 x 399 - 145 / 254 + 399 x 253
=253 x 399 + 399 - 145 / 254 + 399 x 253
= 253 x 399 + ( 399 - 145) / 254 + 399 x 253
= 253 x 399 + 254 / 254 + 399 x 253
= 1
b. 1997 x 1996 - 995 / 1995 x 1997 + 1002 = 1997 x ( 1995 + 1 ) - 995 / 1995 x 1997 + 1002
= 1997 x 1995 + 1997 - 1995 / 1995 x 1997 + 1002
= 1997 x 1995 + ( 1997 - 995 ) / 1995 x1997 + 1002
= 1997 x1995 + 1002 / 1995 x 1997 + 1002
= 1
c. 5932 + 6001 x 5031 / 5932 x 6001 - 69 = 5932 + 6001 x 5031 / ( 5031 + 1 ) x 6001 - 69
= 5932 + 6001 x 5031 / 5031 x 6001 + 6001 - 69
= 5932 + 6001 x 5031 / 5031 x 6001 + ( 6001 - 69 )
= 5932 + 6001 x 5031 / 5031 x 6001 + 5932
= 1
d. 1995 x 1997 - 1 / 1996 x 1995 + 1994 = 1995 x ( 1996 + 1 ) - 1 / 1996 x 1995 + 1994
= 1995 x1996 + 1995 - 1 / 1996 x 1995 + 1994
= 1995 x 1996 + ( 1995 - 1 ) / 1996 x 1995 + 1994
= 1995 x 1996 + 1994 / 1996 x 1995 + 1994
= 1
\(\frac{254.399-145}{254+399.253}=\frac{\left(253+1\right).399-145}{254+399.253}\)
\(=\frac{253.399+1.399-145}{254+399.253}\)
\(=\frac{253.399+254}{254+399.253}\)
\(=1.\)
\(\frac{5932+6001.5931}{5932.6001-69}=\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\frac{5932+6001.5931}{5931.6001+1.6001-69}\)
\(=\frac{5932+6001.5931}{5931.6001+5932}\)
\(=1.\)
Câu a sai đề (sửa đề lại rồi làm tương tự câu b
Câu b
\(\frac{5932+6001\times5931}{5932\times6001-69}=\frac{5932+6001\times5931}{\left(5931+1\right)\times6001-69}=\frac{5932+6001\times5931}{5931\times6001+6001-69}=\frac{5932+6001\times5931}{5932\times6001+5932}=1\)
\(254.399-\frac{145}{254+399.255}=101346-\frac{145}{101999}=1012031170\)
\(5932+6001.\frac{5931}{35597863}=1\)
\(\frac{254.399-145}{254+399.253}=\frac{\left(253+1\right).399-145}{254+399.253}=\frac{253.399+\left(399-145\right)}{254+399.253}\)
\(=\frac{253.399+254}{254+399.253}=1\)
\(\frac{254\times399-145}{254+399\times253}\)
\(=\frac{253\times399+399-145}{254+399\times253}\)
\(=\frac{253\times399+254}{254+399\times253}\)
\(=1\)
254 × 399 - 145 / 254 + 399 × 253
= (253 + 1) × 399 - 145 / 254 + 399 × 253
= 253 × 399 + (399 - 145) / 254 + 399 × 253
= 253 × 399 + 254 / 254 + 399 × 253
= 1
\(\frac{254x399-145}{254+399x253}\)
\(=\frac{253x399+399-145}{254+399x253}\)
\(=\frac{253x399+254}{254+399x253}=1\)
(253+1)x399-145/254+399x253
=253x399+(399-145)/254+399x253
=253x399+254/254+399x253
=1
Ta có:
\(\frac{254X399-145}{254+399X253}\)
\(=\frac{\left(253+1\right)X399-154}{254+399X253}\)
\(=\frac{253X399+399-154}{254+399X253}\)
\(=\frac{253X399+254}{253X399+254}\)
\(=1\)
a) \(\frac{254.399-145}{254+399.253}\)
\(=\frac{253.399+399-145}{254+399.253}\)
\(=\frac{253.399+254}{254+399.253}\)
\(=1\)
b) tự làm