Cho 1 < a< b+c< a+1 và b<c.
Chứng minh b<a
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0=>\frac{ab+bc+ac}{abc}=0=>ab+bc+ac=0.abc=0\)
Mà \(a+b+c=1=>\left(a+b+c\right)^2=1=>a^2+b^2+c^2+2ab+2bc+2ac=1\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=1=>a^2+b^2+c^2=1-0=1\) (vì ab+bc+ac=0)
\(b,S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3\)
\(=2014.\frac{1}{2014}-3=1-3=-2\)
Vậy.....................
1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)
Tương tự : \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)
\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)
1. Câu hỏi của Nguyễn Thị Hồng Nhung - Toán lớp 7 - Học toán với OnlineMath
1)
Ta có :
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\frac{1}{c}:\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{1}{c}.\frac{2}{1}=\frac{\left(a+b\right)}{ab}\)
\(\Leftrightarrow\frac{2}{c}=\frac{\left(a+b\right)}{ab}\)
\(\Leftrightarrow2ab=ac+bc\) (1)
Lại có :
\(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow2ab=ac+bc\) (2)
Từ (1) và (2) :
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Ta có :
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{10}\)
\(\Rightarrow2017\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2017.\frac{1}{10}\)
\(\Rightarrow\frac{2017}{a+b}+\frac{2017}{b+c}+\frac{2017}{c+a}=201,7\)
Mà \(2017=a+b+c\)nên :
\(\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=201,7\)
\(\Rightarrow\left(\frac{a+b}{a+b}+\frac{c}{a+b}\right)+\left(\frac{b+c}{b+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{b}{a+c}\right)=201,7\)
\(3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7\)
\(\Leftrightarrow M=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=201,7-3\)
\(\Leftrightarrow M=198,7\)
Vậy ...
Lời giải:
$(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})=2007.90$
$\Rightarrow \frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{c+a}=180630$
$\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}=180630$
$\Rightarrow M+1+1+1=180630$
$\Rightarrow M =180627$
Mình cần gấp bn nào xong trước mình hs cho
Ta có: \(b< c\Rightarrow b-c< 0\)
Kết hợp với \(b+c< a+\)
\(\Rightarrow\left(b-c\right)+\left(b+c\right)< 0+\left(a+1\right)\)
\(\Rightarrow2b< a+1\)
Lại có: \(1< a\Rightarrow a+1< 2a\)
Suy ra \(2b< a+1< 2a\Rightarrow2b< 2a\)
\(\Rightarrow b< a\)(đpcm)