Tính \(\frac{1}{a}\)- \(\frac{1}{b}\) biết rằng a;b\(\ne\)0 và a-b=ab
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S
11 tháng 9 2017
\(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)
\(=1+\frac{1}{199}+1+\frac{2}{198}+...+\frac{199}{1}+1-199\)
\(=200+\frac{200}{2}+...+\frac{200}{199}-199\)
\(=1+\frac{200}{2}+...+\frac{200}{199}\)
\(=200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{200}\)
SY
15 tháng 3 2016
B= \(\frac{1}{199}\) + \(\frac{2}{198}\) + ... + \(\frac{198}{2}\) + \(\frac{199}{1}\)
B= ( \(\frac{1}{199}\) + 1) + ( \(\frac{2}{198}\) +1) +...+ ( \(\frac{198}{2}\) +1) +1 ( Mình tách 199 ra thành 199 số hạng rồi cộng thêm vào mỗi phân số)
B= \(\frac{200}{199}\) + \(\frac{200}{198}\) + \(\frac{200}{197}\) +...+\(\frac{200}{2}\)
B= 200( \(\frac{1}{199}\) + \(\frac{1}{198}\) +...+ \(\frac{1}{2}\) )
B= 200 ( \(\frac{1}{2}\) + \(\frac{1}{3}\) +...+ \(\frac{1}{198}\) + \(\frac{1}{199}\) ) = 200 A
Ta thấy A=1A, B=200A Suy ra \(\frac{A}{B}\) = \(\frac{1}{200}\)
1 tháng 3 2018
299A=\(\frac{299}{1\cdot300}+\frac{299}{2\cdot301}+...+\frac{299}{101\cdot400}\)
299A=\(1-\frac{1}{300}+\frac{1}{300}-\frac{1}{301}-...-\frac{1}{101}+\frac{1}{101}-\frac{1}{400}\)
299A=\(1-\frac{1}{400}\)
299A=\(\frac{399}{400}\)
A=\(\frac{399}{400}:299\)
A=\(\frac{119310}{400}\)
tương tự tính câu B
1 tháng 3 2018
Ta có: \(A=\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)
\(\Rightarrow A=\frac{1}{399}.\left(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}\right)\)
\(\Rightarrow A=\frac{1}{299.}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(\Rightarrow A=\frac{1}{299}.\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+..+\frac{1}{401}\right)\right]\)
Mặt khác \(B=\frac{1}{1.102}+\frac{1}{2.103}+...+\frac{1}{299.400}\)
\(\Rightarrow B=\frac{1}{101}.\left(1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)
\(\Rightarrow B=\frac{1}{101}.\left[\left(1+\frac{1}{2}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{203}+...+\frac{1}{400}\right)\right]\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{299}.\left[\left(1+\frac{1}{2}+..+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}{\frac{1}{101}.\left[\left(1+\frac{1}{2}+....+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]}\)
\(=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)
DD
0
2 tháng 6 2018
Bài 1 :
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\left(1\right)\)
\(B=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)\(>\frac{1}{10}+\frac{1}{100}.90=1\left(2\right)\)
Từ (1) và ( 2) ta có \(A< 1\) \(B>1\)NÊN \(A< B\)
Bài 2:
\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\frac{\left(a+b+c\right)-\left(b+c\right)}{b+c}+\)\(\frac{\left(a+b+c\right)-\left(c+a\right)}{c+a}\)\(+\frac{\left(a+b+c\right)-\left(a+b\right)}{a+b}\)
\(=\frac{7-\left(b+c\right)}{b+c}+\frac{7-\left(c+a\right)}{c+a}+\frac{7-\left(a+b\right)}{a+b}\)
\(=7.\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(=7.\frac{7}{10}-3\)\(=\frac{49}{10}-3=\frac{19}{10}\)
\(S=\frac{19}{10}>\frac{19}{11}=1\frac{8}{11}\)
Chúc bạn học tốt ( -_- )
2 tháng 6 2018
Bài 1:
ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}< 1\)
\(\Rightarrow A< 1\)(1)
ta có: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) ( có 90 số 1/100)
\(=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{10}+\frac{9}{10}=1\)
\(\Rightarrow B>1\)(2)
Từ (1);(2) => A<B
22 tháng 2 2020
\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
\(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ca+x^2\)
\(=4x^2-2ax-2bc-2cx+ab+bc+ca\)
\(=4x^2-2\left(a+b+c\right)x+ab+bc+ca\)
với \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\Rightarrow2x=a+b+c\)
\(\Rightarrow M=\left(a+b+c\right)^2-\left(a+b+c\right)^2+ab+bc+ca\)
\(=ab+bc+ca\)
\(\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}=\frac{-ab}{ab}=-1\)
\(\Rightarrow\frac{1}{a}-\frac{1}{b}=-1\)