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Đặt :
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+............+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+........+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{3n+2}\)
@Akai Haruma em không hiểu tại sao bài kia chị lại tick cho bạn đó ạ,đề nói chứng minh,mak bạn đó đã làm hết đâu:
\(VT=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{3n-1}+\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{6}-\dfrac{1}{9n+6}\)
\(VT=\dfrac{9n+6}{54n+36}-\dfrac{6}{54n+36}\)
\(VT=\dfrac{9n+6-6}{54n+36}=\dfrac{9n}{54n+36}=\dfrac{9n}{9\left(6n+4\right)}=\dfrac{n}{6n+4}=VP\left(đpcm\right)\)
Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)
\(3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\)
\(=\dfrac{1}{2}-\dfrac{1}{98}\)
\(=\dfrac{24}{49}\)
\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
Sửa đề
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{6n+4}\right)=\dfrac{1}{3}\left(\dfrac{3n}{6n+4}\right)=\dfrac{3n}{18n+12}=\dfrac{3n}{3\left(6n+4\right)}=\dfrac{n}{6n+4}\)
`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`
`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`
`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`
`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`
`3x-5(1/5-1/50)=21/10`
`3x-5*9/50=21/10`
`3x-9/10=21/10`
`3x=21/10+9/10`
`3x=3`
`x=1`
a: \(=\dfrac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\dfrac{1}{2}\)
b: \(=\dfrac{12-15}{20}\cdot\left(\dfrac{10-6}{30}\right)^2\)
\(=\dfrac{-3}{20}\cdot\left(\dfrac{2}{15}\right)^2=\dfrac{-3}{20}\cdot\dfrac{4}{225}=-\dfrac{1}{375}\)
c: \(=3\cdot\dfrac{2}{5}+2:\dfrac{1}{4}=1.2+8=9.2\)
C= \(\dfrac{1}{100}-\)(\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{98.99}\)+\(\dfrac{1}{99.100}\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
=\(\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
= \(\dfrac{1}{100}-\dfrac{99}{100}\)
=\(\dfrac{-98}{100}=-\dfrac{49}{50}\)
Ta có:
\(=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+......+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)
sau khi giản ước ta được như sau:
=\(\dfrac{1}{100}-1\)=\(\dfrac{-99}{100}\)
A = \(\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\)\(\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\)\(\dfrac{1}{72}\)
= \(1-1+\dfrac{7}{12}\)
=\(\dfrac{1}{72}\)
A = ( 1/3+3/5+1/15)-(3/4+2/9+1/36)+1/72
= 1-1+1/72
= 1/72
3.đặt biểu thức trên là A
ta có:
2A=2.(\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{2015.2017}\))
=>2A=\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+\(\dfrac{2}{7.9}\)+....+\(\dfrac{2}{2015.2017}\)
=>2A=\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+....\(\dfrac{1}{2015}\)-\(\dfrac{1}{2017}\)
=>2A=\(\dfrac{1}{3}\)-\(\dfrac{1}{2017}\)=\(\dfrac{2014}{6051}\)
=>A=\(\dfrac{2014}{6051}\):2=\(\dfrac{1007}{6051}\)
Đề sai nhá
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{90.93}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{90.93}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{90}-\dfrac{1}{93}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{93}\right)\)
\(=\dfrac{91}{558}\)