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\(=\dfrac{5931+6001\cdot5931+1}{5931\cdot6001+6001-69}\)
\(=\dfrac{5931\cdot6001+5932}{5931\cdot6001+5932}=1\)
\(\dfrac{5932+6001x5931}{5932x600-69}=\dfrac{5932+6001x\left(5932-1\right)}{5932x6001-69}=\dfrac{5932+6001x5932-6001}{5932x6001-69}=\dfrac{5932x6001-\left(6001-5932\right)}{5932x6001-69}=\dfrac{5932x6001-69}{5932x6001-69}=1\)
Bài 2:
a) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2018}\right)\))
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1}{2018}\)
Bài 1 thì mk ko bk, xl bn nha!
a )\(\frac{13}{15}+\frac{4}{7}-\frac{101}{105}\)
\(=\frac{91}{105}+\frac{60}{105}-\frac{101}{105}\)
\(=\frac{50}{105}\)
\(=\frac{10}{21}\)
b ) \(\frac{2}{5}+\frac{3}{5}\cdot\frac{4}{5}\)
\(=\frac{2}{5}+\frac{12}{25}\)
\(=\frac{10}{25}+\frac{12}{25}\)
\(=\frac{22}{25}\)
c )\(\frac{254\cdot399-145}{254+399\cdot253}\)
\(=\frac{253\cdot399+399-145}{254+399\cdot253}\)
\(=\frac{253\cdot399+254}{254+399\cdot253}\)
\(=1\)
d )\(\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)
\(=\frac{5932+6001\cdot5931}{5931\cdot6001+6001-69}\)
\(=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)
\(=1\)
e )\(\frac{1}{5}\div\frac{2}{7}\)
\(=\frac{1}{5}\cdot\frac{7}{2}\)
\(=\frac{7}{10}\)
hok tốt nha
\(\frac{5932+6001\times5931}{5932\times6001-69}=\frac{5932+6001\times5931}{5931\times6001+6001-69}=\frac{5932+6001\times5931}{5931\times6001+5932}=1\)
a, 35x 34+35x 38 + 65x 75+ 65x 45=10437
b, 36x 28 + 36x 82 + 64x 69 + 64 x41=11000
c, 254x 399 - 145/ 254+ 399 x 253=202292,4291
Viết lại:
\(\frac{5931.6001+5932}{5932.6001-69}=\frac{5931.6001+5932}{5931.6001+6001-69}=\frac{5931.6001+5932}{5931.6001+5932}=1\)
\(A=\frac{254.399-145}{254+399.253}\)
\(=\frac{\left(253+1\right).399-145}{254+399.253}\)
\(=\frac{253.399+399-145}{254+399.253}\)
\(=\frac{253.399+254}{254+399.253}=1\)
mk chỉnh đề
\(B=\frac{5932+6001.5931}{5932.6001-69}\)
\(=\frac{5932+6001.5931}{\left(5931+1\right).6001-69}\)
\(=\frac{5932+6001.5931}{5931.6001+5932}=1\)