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\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{99}{100}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{9.11}{10^2}\)
( đoạn này bạn rút )
\(=\frac{1}{2}.\frac{11}{10}=\frac{22}{20}\)
(mik cx ko chắc, có j sai ib mik sẽ sửa ạ)
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{99}{100}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot2}+\frac{3\cdot5}{4\cdot2}+...+\frac{9\cdot11}{10\cdot10}\)
\(=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{99}{100}=\frac{1\cdot3}{2\cdot2}=\frac{2\cdot4}{3\cdot3}=\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{9\cdot11}{10\cdot10}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)
= ( 1 x 3 / 2 x 2 ) . ( 2 x 4 / 3 x 3 ) . ( 3 x 5 / 4 x 4 ) . . . ( 6 x 8 / 7 x 7 ) ( mk viết thế cho rõ nhìn khi lm bỏ ngoặc cx dc )
= 1 x 3 x 2 x 4 x 3 x 4 x ... x 6 x 8 / 2 x 2 x 3 x 3 x 4 x 4 x ... x 7 x 7
= 1 x 8 / 2 x 7
= 8 / 14
= 4/7
Tk nha !!
Ta có :
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}........\frac{48}{49}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.......\frac{6.8}{7.7}\)
\(=\frac{\left(1.2.3.....6\right).\left(3.4.5.....8\right)}{\left(2.3.4......7\right).\left(2.3.4.....7\right)}\)
\(=\frac{1.2.3.....6}{2.3.4.....7}.\frac{3.4.5......8}{2.3.4......7}\)
\(=\frac{1}{7}.4\)
\(=\frac{4}{7}\)
Đặt\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)
Vì\(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)\(>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(mỗi cái trong ngoặc là một trăm phân số)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\left(\frac{1}{200}\right).100+\left(\frac{1}{300}\right).100\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow A>\frac{5}{6}\)
Mà 5/6>2/3=>A>2/3
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
Đặt A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
Vì \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{103}+.....\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}\right)\)
Tự làm tiếp nhé !!!
P=3/4.8/9.15/16...99/100
P = 3.8.15...99 / 4.9.16...100
P = 1.3.2.4.3.5...9.11 / 2.2.3.3.4.4...10.10
P = (1.2.3.4...9) . (1.2.3.4...9) / (2.3.4...10) . (2.3.4...10)
P = 1.11 / 10.2
P = 11 / 20