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Đặt \(A=1.3.5.7...99\)
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)
Ta có:
\(A=1.3.5.7...99\)
\(\Rightarrow A=\dfrac{\left(1.3.5.7...99\right)\left(2.4.6.8...100\right)}{2.4.6.8...100}\)
\(\Rightarrow A=\dfrac{1.2.3.4...100}{2.4.6.8...100}\)
\(\Rightarrow A=\dfrac{1.2.3.4...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)
\(\Rightarrow A=\dfrac{\left(1.2.3.4...50\right)\left(51.52.53...100\right)}{\left(1.2.3.4...50\right)\left(2.2.2.2...2\right)}\)
\(\Rightarrow A=\dfrac{51.52.53.54...100}{2.2.2.2...2}\)
\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}....\dfrac{100}{2}\)
\(\Rightarrow A=B\)
Vậy \(1.3.5.7...99=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\) (Đpcm)
VT: 1.3.5.7....99=\(\dfrac{(1.3.5.7.....99).\left(2.4.6....100\right)}{2.4.6....100}\)
\(=\dfrac{\left(1.3.5.7.....99\right)\left(2.4.6.....100\right)}{1.2.2.2.2.3.....2.50}\)\(=\dfrac{\left(1.2.3.4.....50\right)\left(51.52.53....100\right)}{\left(1.2.3.4......50\right)\left(2.2.2.2.2....2\right)}\)
\(=\dfrac{51.52.53......100}{2.2.2.2.....2}=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}......\dfrac{100}{2}=VP\left(đpcm\right)\)
Lời giải:
\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)
\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{4}-...-2\cdot\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{50}\)
\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Rightarrow A=B\)
tớ giải chi tiết hơn nhá:
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=(\(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
A=\(\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
A=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
Vậy A=B
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
b)Ta có :
\(A=1.3.5...........99\)
\(\Rightarrow A=\dfrac{\left(1.3.7.9.............99\right)\left(2.4.6.8........100\right)}{2.4.6.8.............100}\)
\(\Rightarrow A=\dfrac{1.2.3.4.............100}{2.4.6.8................100}\)
\(\Rightarrow A=\dfrac{1.2.3.4..................100}{\left(2.1\right)\left(2.2\right)...............\left(2.50\right)}\)
\(\Rightarrow A=\dfrac{51.52.53...........................100}{2.2.2.2.............................2}\)
\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.............\dfrac{100}{2}\)
\(\Rightarrow A=D\)
~ Chúc bn học tốt ~
mk hỏi này sao mà 1.2.3.4.....100/(2.1).(2.2)...(2.50)lại =51.52.53..........100/2.2.2........2