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\(a.\left[x\cdot\left(x+1\right)\right]:2=136\\ x\cdot\left(x+1\right)=136\cdot2\\ x\cdot\left(x+1\right)=2\cdot2\cdot2\cdot17\cdot2\\ x\cdot\left(x+1\right)=\left(2\cdot2\cdot2\cdot2\right)\cdot17\\ x\cdot\left(x+1\right)=16\cdot17\\ =>x=16\\ b.\left[x\cdot\left(x+1\right)\right]:2=300\\ x\cdot\left(x+1\right)=3\cdot2\cdot2\cdot5\cdot5\cdot2\\ x\cdot\left(x+1\right)=\left(3\cdot2\cdot2\cdot2\right)\cdot\left(5\cdot5\right)\\ x\cdot\left(x+1\right)=24\cdot25\\ =>x=24\\ c.\left[x\cdot\left(x+1\right)\right]:2=561\\ x\cdot\left(x+1\right)=2\cdot3\cdot11\cdot17\\ x\cdot\left(x+1\right)=\left(17\cdot2\right)\cdot\left(3\cdot11\right)\\ x\cdot\left(x+1\right)=34\cdot33\\ =>x=33\)
A= 1/3+1/6+1/10+...+1/561
= 2. (1/6+1/12+1/20+...+1/1122)
= 2. [1/(2.3) + 1/(3.4) + 1/(4.5) +...+1/(33.34)]
= 2. ( 1/2 - 1/3 +1/3 - 1/4 + 1/4 - 1/5 +...+ 1/33 - 1/34 )
=2. (1/2 - 1/34)
=2. 8/17
=16/17
Vì 16/17 > 16/18 = 8/9 -> A > 8/9
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{561}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{1122}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{33.34}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{33.34}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{33}-\frac{1}{34}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{34}\right)\)
\(A=2.\left(\frac{17-1}{34}\right)\)
\(A=2.\frac{8}{17}\)
\(A=\frac{16}{17}>\frac{16}{18}=\frac{8}{9}\)
\(\Rightarrow A>\frac{8}{9}\)
đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)
Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)
\(=3-\frac{3}{20}=\frac{57}{20}\)
\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)
Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A
\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)
\(A=\frac{1}{2}.\frac{19}{20}\)
\(A=\frac{19}{40}\)
\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)
\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)
\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)
\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(B=\left[3.\left(\frac{19}{20}\right)\right]\)
\(B=\frac{57}{20}\)
Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)
\(=-\frac{95}{40}=-\frac{19}{8}\)
Nếu đúng thì k nha
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.......+\frac{1}{100^2}<\frac{1}{2}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)<\(\frac{1}{0.2}+\frac{1}{2.4}+\frac{1}{4.6}+.......+\frac{1}{98.100}\)
\(S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{49}{100}<\frac{1}{2}\)
Vậy \(\frac{49}{100}<\frac{1}{2}\)
Ta có 1/22<1/2*3
1/42<1/3*4
. . .
1/1002<1/99*100
=> S<1/2*3+1/3*4+...+1/99*100
=> S<1/2-1/3+1/3-1/4+...+1/99-1/100
=>S<1/2-1/100
=>S<49/100
Mà 49/100<1/2
=>S<1/2
1)
Gọi hiệu đó là a - b = c
=> a = b + c
Tổng theo đề bài là : a + b + c
Thay a = b + c ta có :
a + b + c = a + a = 2a chia hết cho 2 ( đpcm )
\(A=\frac{5^{60}+1}{5^{61}+1}\)
\(5A=\frac{5(5^{60}+1)}{5^{61}+1}=\frac{5^{61}+5}{5^{61}+1}=\frac{5^{61}+1+4}{5^{61}+1}=1+\frac{4}{5^{61}+1}\) \((1)\)
\(B=\frac{5^{61}+1}{5^{62}+1}\)
\(5B=\frac{5(5^{61})+1}{5^{62}+1}=\frac{5^{62}+5}{5^{62}+1}=\frac{5^{62}+1+4}{5^{62}+1}=1+\frac{4}{5^{62}+1}\) \((2)\)
Từ 1 và 2 \(\Rightarrow1+\frac{4}{5^{61}+1}>1+\frac{4}{5^{62}+1}\)
\(\Rightarrow5A>5B\)
Hay \(A>B\)
Vậy : ...