tính giá trị biểu thức : \(\dfrac{4}{1.2}\)+\(\dfrac{4}{2.3}\)+\(\dfrac{4}{3.4}\)+ ... + \(\dfrac{4}{2014.2015}\)
Đề: Tính giá trị biểu thức
\(A=\dfrac{4}{1.2}+\dfrac{4}{2.3}+\dfrac{4}{3.4}+...+\dfrac{4}{2014.2015}\)
Mình k ghi lại đề nhé!~~
\(A=\dfrac{4}{1}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)
\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2015}\right)\)
\(A=\dfrac{4}{1}.\left(\dfrac{2015-1}{2015}\right)\)
\(A=\dfrac{4}{1}.\dfrac{2014}{2015}\)
\(A=3,998014888\)
\(A\approx4\)
Tính giá trị biểu thức : \(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{99}{100}\)
\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)
\(P=1+0+0+....+0+\dfrac{-1}{100}\)
\(P=1+\dfrac{-1}{100}\)
\(P=\dfrac{99}{100}\)
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100
=99/100
tính giá trị biểu thức : A= 4/1.2+4/2.3+4/3.4+...4/2014.2015
\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)
Tính giá trị các biểu thức :
\(A=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\) \(B=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
a) A = \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
A = \(\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}.\dfrac{4.4}{4.5}\)
A = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)= \(\dfrac{1}{5}\)
b) B = \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
B = \(\dfrac{2.3.4.5}{1.2.3.4}.\dfrac{2.3.4.5}{3.4.5.6}\)= \(\dfrac{5}{3}\)
Tính giá trị biểu thức sau:
A = \(\dfrac{7}{1.2}\) + \(\dfrac{7}{2.3}\) + \(\dfrac{7}{3.4}\) +... + \(\dfrac{7}{2011.2012}\)
Giúp mik nhanh câu này
\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)
\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)
\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)
\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)
Tính hợp lí:
\(\dfrac{-4}{1.2}+\dfrac{-4}{2.3}+\dfrac{-4}{3.4}+...+\dfrac{-4}{97.98}+\dfrac{-4}{98.99}\)
Tính giá trị của biểu thực:
P=\(\dfrac{5}{1.2}\)+\(\dfrac{5}{2.3}\)+\(\dfrac{5}{3.4}\)+...+\(\dfrac{5}{2023.2024}\)
=5(1-1/2+1/2-1/3+...+1/2023-1/2024)
=5*2023/2024
=10115/2024
tính các tổng sau bằng cách hợp lí
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2021.2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+......+\dfrac{4}{107.111}\)
\(S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+.....+\dfrac{1}{60}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)
Bài 1 : Tìm 2 số biết hiệu của chúng bằng 5 và 50% số lớn = 1 nửa số bé.
Bài 2 : tính giá trị biểu thức: A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
Bài 3 : Tìm x
a , \(\dfrac{x}{3}-\dfrac{1}{8}=\dfrac{5}{8}\)
bài 2:
\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)
bài 3:
\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)
\(=>x=3\)