Bài 4 Tính
a)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2020}-1\right)\left(\frac{1}{2021}-1\right)\)
K
Khách
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2 tháng 6 2017
\(A=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{100}\right)\)
\(A=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{101}{100}\)
\(A=\frac{101}{2}\)
VT
30 tháng 4 2017
A = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{101}{100}\)
A = \(\frac{101}{2}\)
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
KT
20 tháng 7 2018
\(A=\left(-2\right)\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{214}\right)\)
\(=2.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{215}{214}=215\)
\(B=\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)....\left(-1\frac{1}{299}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{300}{299}=\frac{300}{2}=150\)
\(C=-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{333333}{424242}\right)\)
\(=-\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=-\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=-\frac{231}{4}.\frac{4}{21}=-11\)
6 tháng 10 2020
a) \(\left(0,25+\frac{3}{4}:1,25+1\frac{1}{3}:2\right)\)\(=\left(\frac{1}{4}+\frac{3}{4}\right):\frac{5}{4}+\frac{4}{3}:2\)\(=\frac{4}{5}+\frac{2}{3}=\frac{22}{15}\)
b) \(2^3+3\left(\frac{-3}{2}\right)^0.\left(\frac{1}{2}\right)^2.4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)\(=8+3.1.\frac{1}{4}.4+\left[4:\frac{1}{2}\right]:8\)
\(=8+3+8:8=\frac{19}{8}\)
13 tháng 4 2020
a) Ta có: \(\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)=\left(4-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)-\left(x-4\right)\right]-2\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3-x+4\right)-6x+4=0\)
\(\Leftrightarrow x-4-6x+4=0\)
\(\Leftrightarrow-5x=0\)
mà -5<0
nên x=0
Vậy: x=0
HT
18 tháng 10 2020
iem chỉ biết làm câu đầu , NHƯNG KO BÍT có ĐUG HAY KO
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2018\cdot2019}{2\cdot3\cdot4\cdot..\cdot2019\cdot2020}\)
\(A=\frac{1}{2020}\)
TA
18 tháng 10 2020
Với \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\) , ta có : \(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}=\frac{1}{2020}\)
Ta có \(7A=\frac{7}{2020}\) , \(9A=\frac{9}{2020}\) , \(1+A=\frac{2021}{2020}\)
\(\frac{1+7A}{1+9A}=\frac{1+\frac{7}{2020}}{1+\frac{9}{2020}}=\frac{\frac{2027}{2020}}{\frac{2029}{2020}}\)
Ta thấy \(\frac{\frac{2027}{2020}}{\frac{2029}{2020}}\)có tử kém mẫu \(\frac{2}{2020}\)đơn vị và không thể rút gọn được nữa .
\(\Rightarrow\frac{1+7A}{1+9A}\)là p/s tối giản.
23 tháng 8 2018
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2021}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right)....\left(-\frac{2020}{2021}\right)\)
\(=\frac{1.2.3...2020}{2.3.4...2021}=\frac{1}{2021}\)
Ta có
(1−1/2)×(1−1/3)×(1−1/4).....×(1−1/2020)×(1−1/2021)(1-1/2)×(1-1/3)×(1-1/4).....×(1-1/2020)×(1-1/2021)
=1/2×2/3×3/4.....×2019/2020×2020/2021=1/2×2/3×3/4.....×2019/2020×2020/2021
=1×2×3×.....×2019×2020/2×3×4×....×2020×2021=1×2×3×.....×2019×2020/2×3×4×....×2020×2021
=1/2021