1.Tính nhanh:
a. 2022 - 542 + 256.352 b. 621 - 769.373 - 1482
c. \(\frac{42^2-10^2}{\left(36,5\right)^2-\left(27,5\right)^2}\) d. S = \(\frac{93^2+83^2}{180}\) - 97.83
2.Tìm 4 số nguyên liên tiếp biết tích của chúng bằng 120
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1
a,
=(202+54).(202-54)+256.352
=37888+256.352
=37888+90112
=128000
b,
=621-769.373-21904
=621-286837-21904
=-308120
c,
42^2-10^2/(36,5)^2-(27,5)^2
=(42-10).(42+10)/(36,5-27,5).(27,5+36,5)
=1664/576=2(8)
1. Tính nhanh :
a) \(202^2-54^2+256.352\)
\(=\left(202-54\right).\left(202+54\right)+256.352\)
\(=148.256+256.352\)
\(=256.\left(148+252\right)=256.400=102400\)
\(R=\frac{43^2-11^2}{\left(36,5\right)^2-\left(27,5\right)^2}\)
\(=\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5-27,5\right)\left(36,5+27,5\right)}\)
\(=\frac{32.54}{9.64}\)
\(=\frac{6}{2}=3\)
Bạn viết sai đề bài rồi
\(S=\frac{97^3+83^3}{180}-97.83\)
\(=\frac{\left(97+83\right)\left(97^2-97.83+83^2\right)}{180}-97.83\)
\(=97^2-97.83+83-97.83\)
\(=\left(97-83\right)^2=14^2=196\)
Trả lời:
\(R=\frac{43^2-11^2}{36,5^2-27,5^2}\)
\(R=\frac{\left(43-11\right).\left(43+11\right)}{\left(36,5-27,5\right).\left(36,5+27,5\right)}\)
\(R=\frac{32.54}{9.64}\)
\(R=3\)
Đề bài sai bạn nhé
\(S=\frac{97^3+83^3}{180}-97.83\)
\(S=\frac{\left(97+83\right).\left(97^2-97.23+83^2\right)}{180}-97.83\)
\(S=97^2-97.83+83^2-97.83\)
\(S=97^2-2.97.83+83^2\)
\(S=\left(97-83\right)^2\)
\(S=14^2\)
\(S=196\)
\(R=\frac{43^2-11^2}{36,5^2-27,5^2}\)
\(R=\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5+27,5\right)\left(36,5-27,5\right)}\)
\(R=\frac{32.54}{64.9}\)
\(R=3\)
\(\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5-27,5\right)\left(36,5+27,5\right)}=\frac{32.54}{9.64}\)
\(=3\)
dùng hằng đẳng thức thứ 3 ta có
(43-11)(43+11) và (36,5-27,5)(36,5+27,5)
Nên =1728 : 576=3
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
a) \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)
\(=\frac{31}{3}+\frac{4}{9}-\frac{26}{3}\)
\(=\left(\frac{31}{3}-\frac{26}{3}\right)+\frac{4}{9}=\frac{5}{3}+\frac{4}{9}=\frac{15}{9}+\frac{4}{9}=\frac{19}{9}\)
b) \(\left[12,\left(1\right)-2,3\left(6\right)\right]:4,\left(21\right)\)
\(=\left[\frac{109}{9}-\frac{71}{30}\right]:\frac{139}{33}\)
\(=-\frac{52}{45}:\frac{139}{33}=-\frac{52}{45}\cdot\frac{33}{139}=-\frac{572}{2085}\)(số xấu quá)
c) \(3\frac{1}{2}\cdot\frac{4}{49}-\left[2,\left(4\right)\cdot2\frac{5}{11}\right]:\frac{-42}{53}\)
\(=\frac{7}{2}\cdot\frac{4}{49}-\left[\frac{22}{9}\cdot\frac{27}{11}\right]\cdot\frac{-53}{42}\)
\(=\frac{2}{7}-6\cdot\left(-\frac{53}{42}\right)=\frac{2}{7}-\left(-\frac{53}{7}\right)=\frac{2}{7}+\frac{53}{7}=\frac{55}{7}\)