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a) \(\frac{158\cdot168-168\cdot58}{110}\)
\(=\frac{\left(158-58\right)\cdot168}{110}\)
\(=\frac{100\cdot168}{110}=\frac{16800}{110}=\frac{1680}{11}\)
b) \(\frac{\left(456.11+912\right)\cdot37}{13\cdot74}\)
\(=\frac{456\cdot11+456\cdot2\cdot37}{13\cdot37\cdot2}\)
\(=\frac{456\cdot\left(11+2\right)\cdot37}{\left(13\cdot2\right)\cdot37}\)
\(=\frac{456\cdot13\cdot37}{26\cdot37}=228\)
c) \(\frac{864\cdot48-432\cdot96}{864\cdot48\cdot432}\)
\(=\frac{432\cdot2\cdot48-432\cdot48\cdot2}{432\cdot2\cdot48\cdot432}=0\)
Đây là những bài lớp 5+ vậy nên không làm theo kiểu lớp 5 được, cố gắng tìm bài đúng lớp nhé bạn (VNNLL cc)
\(=\left(45\cdot128-45\cdot128\right)\cdot\left(45\cdot46+47\cdot48\right)\cdot\left(51\cdot52-49\cdot48\right)\cdot\left(1995\cdot1996+1997\cdot1998\right)\)
=0*A
=0
Số đó là :
864 : \(\frac{1}{4}\)= 3456
\(\frac{3}{4}\)số đó là :
3456 x \(\frac{3}{4}\)= 2592
Đáp số : 2592
\(\frac{3}{4}\)số đó là :
864 x \(\frac{3}{4}\)= 648
Đáp sô : 648
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
\(B1\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)
\(=1-\frac{1}{39}\)
\(=\frac{38}{39}\)
\(B2\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}\)
\(=\frac{24}{100}\)
\(=\frac{6}{25}\)
Bài 1 :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{370}{741}\)
4524 - (864 - 999) - (36 + 3999)
= 4524 - 864 + 999 - 36 - 3999
= (4524 - 864) + (999 - 3999) - 36
= 3660 - 3000 - 36
= 660 - 36
= 624.
(Tíck cho mìk vs nha!)
864.48-432.96 = 432.2.48-432.96=432.96-432.96=...=0
.....=> \(\frac{864.48-432.96}{2003.2004.2017}=0\)
Sai đề!