Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
bình phương 2 vế của 1/a + 1/b +1/c =2 ta đk:
1/a^2 +1/b^2 + 1/c^2 + 2 x (a+b+c) / abc =4
1/a^2 + 1/b^2 + 1/c^2 +2 =4
=> 1/a^2 + 1/b^2 + 1/c^2 =2
olm đã có
a: Ta có: \(2x^3-5x^2+8x-3=0\)
\(\Leftrightarrow2x^3-x^2-4x^2+2x+6x-3=0\)
=>2x-1=0
hay x=1/2
cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)
\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)
\(=x-1=2013-1=2012\)
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(=>Q=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=>Q=\left(\frac{a+b+c}{b+c}\right)+\left(\frac{a+b+c}{a+c}\right)+\left(\frac{a+b+c}{a+b}\right)-3\)
\(=>Q=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(=>Q=259.15-3=3882\)
Vậy Q=3882
\(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{259-\left(b+c\right)}{b+c}+\frac{259-\left(a+c\right)}{a+c}+\frac{259-\left(a+b\right)}{a+b}\)
\(=259.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)+\left[\frac{-\left(b+c\right)}{b+c}+\frac{-\left(a+c\right)}{a+c}+\frac{-\left(a+b\right)}{a+b}\right]\)
tới đây tự làm tiếp
Bài 3:
Do a và b đều không chia hết cho 3 nhưng khi chia cho 3 thì có cùng số dư nên\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=3n+1\\b=3m+1\end{matrix}\right.\\\left\{{}\begin{matrix}a=3n+2\\b=3m+2\end{matrix}\right.\end{matrix}\right.\)
TH1:\(\left\{{}\begin{matrix}a=3n+1\\b=3m+1\end{matrix}\right.\)
\(\Rightarrow ab-1=\left(3n+1\right)\left(3m+1\right)-1\)
\(\Rightarrow ab-1=9nm+3m+3n+1-1=9nm+3m+3n⋮3\) nên là bội của 3 (đpcm)
TH2:\(\left\{{}\begin{matrix}a=3n+2\\b=3m+2\end{matrix}\right.\)
\(\Rightarrow ab-1=\left(3n+2\right)\left(3m+2\right)-1\)
\(\Rightarrow ab-1=9nm+6m+6n+4-1=9nm+6m+6n+3⋮3\) nên là bội của 3 (đpcm)
Vậy ....
Bài 2:
\(B=\frac{1}{2010.2009}-\frac{1}{2009.2008}-\frac{1}{2008.2007}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{2010.2009}-\left(\frac{1}{2009.2008}+\frac{1}{2008.2007}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
Đặt A=\(\frac{1}{2009.2008}+\frac{1}{2008.2007}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(\Rightarrow A=\frac{2009-2008}{2009.2008}+\frac{2008-2007}{2008.2007}+...+\frac{3-2}{3.2}+\frac{2-1}{2.1}\)
\(\Rightarrow A=\frac{2-1}{2.1}+\frac{3-2}{3.2}+...+\frac{2008-2007}{2008.2007}+\frac{2009-2008}{2009.2008}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008}+\frac{1}{2008}-\frac{1}{2009}\)
\(\Rightarrow A=1-\frac{1}{2009}\)
\(\Rightarrow B=\frac{1}{2010.2009}-A=\frac{1}{2010.2009}-\left(1-\frac{1}{2009}\right)\)
\(\Rightarrow B=\frac{1}{2010.2009}+\frac{1}{2009}-1=\frac{2011}{2010.2009}-1\)
Đặt x=a+b+c(x>3)
Ta có \(\left(x-6\right)^2\ge0\)(dấu '=' xảy ra khi x=6 hay a+b+c=6)\(\Leftrightarrow x^2-12x+36\ge0\Leftrightarrow x^2\ge12x-36\Leftrightarrow x^2\ge12\left(x-3\right)\Leftrightarrow\frac{x^2}{x-3}\ge12\)(1)
Áp dụng bđt \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)(dấu '=' xảy ra khi \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\))
Ta có \(\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}\ge\frac{\left(a+b+c\right)^2}{a+b+c-3}=\frac{x^2}{x-3}\)(2)
Từ (1) và (2)\(\Rightarrow\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}\ge12\)(đpcm)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}\frac{a}{b-1}=\frac{b}{c-1}=\frac{c}{a-1}\\a+b+c=6\end{matrix}\right.\)\(\Leftrightarrow a=b=c=2\)