tính giá trị của các biểu thức
a) A= \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
b) B= \(\frac{17}{1.3.5}+\frac{17}{3.5.7}+...+\frac{17}{47.49.51}\)
Tính :
A = \(\frac{17}{1.3.5}\) + \(\frac{17}{3.5.7}\) + \(\frac{17}{5.7.9}\) + ... + \(\frac{17}{47.49.51}\)
Ủng hộ 2 tick cho !!!
\(=\frac{1}{4}.\left(\frac{17.4}{1.3.5}+\frac{17.4}{3.5.7}+\frac{17.4}{5.7.9}+...+\frac{17.4}{47.49.51}\right)\)
\(=\frac{17}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)
\(=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}.\frac{832}{2499}=\frac{208}{147}\)
tính giá trị của các biểu thức sau:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{98.99.100}\)
2A=2(1/1.2.3+1/2.3.4+...+1/98.99.100)
2A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=4949/9900
A=4949/9900:2
A=4949/19800
Vậy A=4949/198000
Lưu Trường An làm đúng rồi
Tính giá trị
\(C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
Tính giá trị biểu thức :\(P=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
Tính:
a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
b) \(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
a) \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{1.2}-\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{9900}\)
\(A=\frac{9898}{19800}.\)
Vậy :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{9898}{19800}:2\)
\(A=\frac{4949}{19800}.\)
a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
A = \(\frac{1}{2}.\frac{4949}{9900}\)
A = \(\frac{4949}{19800}\)
Tính C = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
cách làm như sau
\(C=\frac{2}{2}.\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}\right]\)
\(C=1\left[\frac{1}{2}-\frac{1}{9900}\right]\)
\(C=\frac{4949}{9900}\)
Chứng minh rằng :
a) A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}<\frac{1}{4}\)
b) B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}<3\)
$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$
Tính\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
Tính giá trị biểu thức: A=\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)....\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right).}\)