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1.Tính D = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/18.19.20 - 3/1.2 - 3/2.3 - 3/4.5 - ...- 3/19.20
Lời giải chi tiết nha,làm ơn giải giùm mình nha !
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19 tháng 7 2017
<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x
2x^2-8x-17-2x^2-2=0
-8x-19=0
x=-19/8
9 tháng 9 2017
a)1+3+5+7+9+...+x=1600
=>[(x-1):2+1].(x+1)/2=1600
=>(1/2.x-1/2+1).(x+1)=1600:1/2
=>(1/2.x-1/2+1).(x+1)=3200
=>(x+1)2.1/2=3200
=>(x+1)2 =3200:1/2
=>(x+1)2=6400
=>x+1=80
=>x=80-1=79
NN
20 tháng 7 2023
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
TD
20 tháng 1 2019
Đặt biểu thức là A
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+.2018.2019\)
\(2A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(2A=\frac{1}{2}-\frac{1}{2018.2019}\)
A= 1/4 - 1/(2018.2019)
Vậy A = ... (tự ghi)
20 tháng 1 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2017.2018.2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2018.2019}\right)\)
Tự làm nốt
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