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a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
Baif: A=\(\frac{10n}{5n-3}=2+\frac{6}{5n-3}\)
để A nguyên thì 5n-3 = Ư(6)={-1;-2;-3;-6;1;2;3;6}
xét từng TH:
- 5n-3=-1=>n=2/5
- 5n-3=-2=>n=1/5
- 5n-3=-3=>n=0
- 5n-3=-6=>n=-3/5
- 5n-3=1=>n=4/5
- 5n-3=2=>n=1
- 5n-3=3=>n=6/5
- 5n-3=6=>n=9/5
b) A= \(\frac{10n}{5n-3}=2+\frac{6}{5n-3}\)
để A lớn nhất thì 5n-3 nhỏ nhất
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)\cdot0\)
\(C=0\)
\(C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)x\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Ta thấy 1/2-1/3-1/6=1/6-1/6=0
\(\Rightarrow C=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right)x0\)
\(\Rightarrow C=0\)
Vậy...............
\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5+3+2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times0\)
\(=0\)
_Chúc bạn học tốt_
\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times\left(\frac{-5+3+2}{12}\right)\)
\(=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right)\times0\)
\(=0\)
_Chúc bạn học tốt_
Câu 1 có vế sau = 0
Câu 2 có vế trước = 0
Một biểu thức nhân với 0 thì = 0, nên:
=> kết quả hai bài đều = 0