Tính giá trị biểu thức A=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\)ta được kết quả A=...
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16 tháng 9 2021
\(A=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\\ B=\dfrac{25}{11}\times\dfrac{13}{12}\times\dfrac{-11}{5}=\dfrac{5\times13\times\left(-1\right)}{1\times12\times1}=\dfrac{-65}{12}\\ C=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\times\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\times\dfrac{-2}{5}=\dfrac{-11}{50}\)
\(B< -1< C< 0< A\\ \Leftrightarrow B< C< A\)
9 tháng 9 2023
a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)
\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)
25 tháng 9
Mới thế đã hai năm trôi qua,câu trả lời từ mọi người vẫn KO XUẤT HIỆN.
Ko biết sau này câu trả lời có xuất hiện hay ko...
17 tháng 11 2023
a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)
b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)
\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)
\(=\dfrac{x-\left(x-5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}\)
c: \(x^3-x^2+2=0\)
=>\(x^3+x^2-2x^2+2=0\)
=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)
=>x+1=0
=>x=-1
Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)
18 tháng 4 2021
a, ĐK : \(x\ne1;2;3;4;5\)
b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)
\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 5 2023
A = \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)
A = \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{99\times100}\)
A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{97}{300}\)
AH
Akai Haruma
Giáo viên
1 tháng 5 2023
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{99.100}$
$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}$
H9
HT.Phong (9A5)
CTVHS
30 tháng 8 2023
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
A=\(1-\dfrac{1}{10}\)
A=\(\dfrac{9}{10}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
A= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
A= \(\dfrac{1}{1}-\dfrac{1}{10}\)
A= \(\dfrac{9}{10}\)