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b) \(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot...\cdot\frac{100^2}{100\cdot101}=\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{1\cdot2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{2\cdot3\cdot4\cdot...\cdot101}=1\cdot\frac{1}{101}=\frac{1}{101}\)
a không biết
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(A=\frac{1.1.2.2.3.3...9.9}{1.2.2.3.3.4...9.10}\)
\(A=\frac{1}{10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-1\right)\)
\(B=\frac{1}{99}-\frac{1}{99}+1\)
\(B=1\)
\(A=\frac{1\cdot1}{1\cdot2}\cdot\frac{2\cdot2}{2\cdot3}\cdot\frac{3\cdot3}{3\cdot4}\cdot\frac{4\cdot4}{4\cdot5}=\frac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot4}\cdot\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)
\(=\frac{1.2.3.....99}{1.2.3.....98}.\frac{1.2.3......99}{2.3.4.5....100}\)
\(=99.\frac{1}{100}\)
\(=\frac{99}{100}\)
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
Xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{2}{2.3}=\frac{1.4}{2.3}\) ; \(1-\frac{2}{3.4}=\frac{2.5}{3.4}\) ; ... ; \(1-\frac{2}{101.102}=\frac{100.103}{101.102}\)
\(\Rightarrow M=\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\cdot\cdot\frac{100.103}{101.102}\)
\(M=\frac{\left(1.2...100\right).\left(4.5...103\right)}{\left(2.3...101\right).\left(3.4...102\right)}=\frac{103}{101.3}=\frac{103}{303}\)
Vậy \(M=\frac{103}{303}\)
1)xE{2;0} 2)abcd=a000+b00+c0+d=a.1000+b.100+c.10+d=(a.1000+b.96+c.8)+(4.b+2.c+d)=8.(a.125+b.12+c)+(d+2.c+4.b). vì 8 chia hết cho 8 =>8.(a.125+b.12+c) chia hết cho 8. Mà d+2.c+4.b chia hết cho 8. =>8.(a.125+b.12+c)+(d+2.c+4.b) chia hết cho 8 hay abcd chia hết cho 8. 3)3.S=1.2.3+2.3.3+3.4.3+...+99.100.3. =>3S=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98) =>3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100. =>3S=99.100.101.=>3s=979902=>S=326634.
2/1.2*2/20.21
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{20.21}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\left(1-\frac{1}{21}\right)=2.\frac{20}{21}=\frac{40}{21}\)