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Ta có:
31/2.32/2.33/2....60/2=31.32......60/2^30
=(31.32.33....60)(1.2.3....30)/2^30(1.2.3...30)
=(1.3.5...59)(2.4.6...60)/(2.4.6...60)=1.3.5...59
=>P=Q
nhớ ****
cái dòng 3, 4 mk ko hiểu sao 2^30.(1.2.3....30) lại bằng 2.4.6...60
\(60!=1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot59\cdot60=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2\cdot4\cdot6\cdot...\cdot58\cdot60\)
\(=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\cdot1\cdot2\cdot3\cdot...\cdot30=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\times30!\)
\(\Rightarrow1\cdot3\cdot5\cdot...\cdot59=\frac{60!}{30!\times2^{30}}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}\)đpcm.
\(\frac{31}{2}\cdot\frac{32}{2}\cdot...\cdot\frac{60}{2}\cdot2\cdot4\cdot...\cdot58\cdot60\)
=31.32.33.34...60.1.2.3.4.5...29.30
=1.2.3.4.5.6.7.8.9.10...60
1.3.5.7...59.2.4.6.8...60
=1.2.3.4.5.6...60
Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}=1\cdot3\cdot5\cdot...\cdot59\)
Câu hỏi của Quỳnh Anh - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo câu 1 2 cách 2 bạn hướng dẫn nhé!
\(1.3.5....59=\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}=\frac{\left(1.2.3.4.5...30\right).31....59.60}{2^{30}.\left(1.2.3...30\right)}=\frac{31.32....60}{2^{30}}=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}\)
Ta có: \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2.2.2...2}=\frac{31.32.33...60}{2^{30}}\)
(30 số 2)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}=\frac{31.32.33...60.1.2.3...30}{\left(2.1\right).\left(2.2\right).\left(2.3\right)...\left(2.30\right)}=\frac{\left(1.3.5...59\right).\left(2.4.6...60\right)}{\left(2.4.6...60\right)}=1.3.5...59\)
\(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2^{30}}\)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}\)
\(=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{\left(1.3.5.7...59\right)\left(2.4.6.8...60\right)}{\left(2.4.6.8...60\right)}\)
\(=1.3.5.7...59\)
Ta có:
\(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=\frac{31.32.33...60}{2^{30}}\)
\(=\frac{\left(31.32.33...60\right).\left(1.2.3...30\right)}{2^{30}.\left(1.2.3...30\right)}\)
\(=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{\left(1.3.5.7...59\right)\left(2.4.6.8...60\right)}{\left(2.4.6.8...60\right)}\)
\(=1.3.5.7...59\)
Vậy \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}=1.3.5.7...59\)
