Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=7\cdot\frac{7}{10}=\frac{49}{10}\)
\(\Leftrightarrow\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{a+c}{a+c}+\frac{b}{a+c}+\frac{b+c}{b+c}+\frac{a}{b+c}=\frac{49}{10}\)
\(3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{49}{10}\Leftrightarrow S=\frac{19}{10}\)
Ta có: \(1\frac{8}{11}=\frac{19}{11}\)
vì 19=19 ,\(\frac{1}{11}< \frac{1}{10}\)nên \(\frac{19}{11}< \frac{19}{10}\)
Vậy \(S>1\frac{8}{11}\)
a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
b) b = a - c => b + c = a
\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)
Bước 2 bạn sai rồi. Vd: \(\frac{1}{3x3}\) đâu bằng hay nhỏ hơn \(\frac{1}{2x3}\)
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 ( -_- )
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
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Rightarrow\frac{bc+ac}{abc}=\frac{ab}{abc}\Rightarrow bc+ac=ab\)
\(\Rightarrow ab-ac-bc=0\Rightarrow a\left(b-c\right)-c\left(b-c\right)=c^2\)
\(\Rightarrow\left(b-c\right)\left(a-c\right)=c^2\Rightarrow\frac{a-c}{c}=\frac{c}{b-c}\)
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
TA có
\(\frac{a}{b}-\frac{a+c}{b+c}=\frac{a\left(b+c\right)}{b\left(b+c\right)}-\frac{b\left(a+c\right)}{b\left(b+c\right)}\)
\(=\frac{ab+ac-ab-bc}{b\left(b+c\right)}=\frac{ac-bc}{b\left(b+c\right)}=\frac{c\left(a-b\right)}{b\left(b+c\right)}\)
vì a>b => a-b > 0 => c(a-b) > 0
=> \(\frac{c\left(a-b\right)}{b\left(b+c\right)}>0\)
\(=>\frac{a}{b}-\frac{a+c}{b+c}>0\)
\(=>\frac{a}{b}>\frac{a+c}{b+c}\)
=> đpcm
b) Ta có a+b < a+b+c ; b+c < a+b+c ; c+a < a+b+c
\(=>\frac{a}{a+b}>\frac{a}{a+b+c};\frac{b}{b+c}>\frac{b}{a+b+c};\frac{c}{c+a}>\frac{c}{a+b+c}\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a+b+c}{a+b+c}=1\) (1)
Lại có
Áp dùng câu a ta có a< a+b ; b< b+c ; c<c+a
=> \(\frac{a}{a+b}< \frac{a+c}{a+b+c};\frac{b}{b+c}< \frac{b+a}{a+b+c};\frac{c}{c+a}< \frac{c+b}{a+b+c}\)
\(=>\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{2\left(a+b+c\right)}{a+b+c}=2\) (2)
Từ (1) và (2) => dpcm
Ta có :
\(S+3=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(=2016\cdot\frac{1}{90}=\frac{112}{5}\)
\(\Rightarrow S=\frac{112}{5}-3=\frac{97}{5}\)