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Lời giải:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow \frac{ab+bc+ac}{abc}=0\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow 2(ab+bc+ac)=0\)
Cộng cả hai vế với \(a^2+b^2+c^2\) thì:
\(a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2\)
\(\Leftrightarrow (a+b+c)^2=a^2+b^2+c^2\)
Do đó ta có đpcm.
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{a}=x\\\dfrac{1}{b}=y\\\dfrac{1}{c}=z\end{matrix}\right.\) \(\Rightarrow x+y+z=0\) \(\Rightarrow z=-\left(x+y\right)\)
Đẳng thức cần chứng minh: \(x^3+y^3+z^3=3xyz\) với \(x+y+z=0\)
Ta có:
\(x^3+y^3+z^3=x^3+y^3-\left(x+y\right)^3=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2-\left(x+y\right)^2\right)=\left(x+y\right)\left(-3xy\right)\)
\(=-\left(x+y\right).3xy=z.3xy=3xyz\)
Vậy \(x^3+y^3+z^3=3xyz\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(\)
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+\left(a-c\right)^2+c^2+2\left(ab-ac-bc\right)}{b^2+\left(b-c\right)^2+c^2+2\left(ab-ac-bc\right)}\)
\(=\dfrac{a^2+a^2-2ac+c^2+c^2+2ab-2ac-2bc}{b^2+b^2-2bc+c^2+c^2+2ab-2ac-2bc}\)
\(=\dfrac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}\)
\(=\dfrac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(a-c+b\right)}=\dfrac{a-c}{b-c}\left(đpcm\right)\)
Lời giải:
Ta có:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow \left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)(a+b+c)=a+b+c\)
\(\Leftrightarrow \frac{a^2}{b+c}+\frac{a(b+c)}{b+c}+\frac{b(c+a)}{c+a}+\frac{b^2}{c+a}+\frac{c(a+b)}{a+b}+\frac{c^2}{a+b}=a+b+c\)
\(\Leftrightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Leftrightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Ta có đpcm.
Ta có a+b+c=0 => b+c=-a => a^2=b^2+2bc+c^2=> a^2-b^2-c^2=2bc
Tương tự ta có : b^2-c^2-a^2=2ca
c^2-a^2-b^2=2ab
=> a^2/2bc+b^2/2ca+c^2/2ab=(a^3+b^3+c^3)/2abc
=>Ta lại có a^3+b^3+c^3=(a+b+c)^3+
Từ giả thiết ta có:
\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^2=a^2\)
\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)
Tương tự:
\(b^2-c^2-a^2=2ca,c^2-a^2-b^2=2ab\)
Từ đây suy ra:
\(A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{2abc}\)
Mặt khác lại có:
\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^3=-a^3\)
\(\Rightarrow b^3+c^3+3bc\left(b+c\right)=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
\(\Rightarrow A=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3}{2}\)
ab−c−ba−c−cb−a=0=>ab−c−ba−c−cb−a=0
=>ab−c=ba−c+cb−a=b2−ab+ac−c2(c−a)(a−b)=>ab−c=ba−c+cb−a=b2−ab+ac−c2(c−a)(a−b)
Nhân cả 2 vế với 1b−c1b−c ta được
a(b−c)2=b2−ab+ac−c2(a−b)(b−c)(c−a)(1)a(b−c)2=b2−ab+ac−c2(a−b)(b−c)(c−a)(1)
Tương tự ta có:
b(c−a)2=c2−bc+bc−a2(a−b)(b−c)(c−a)(2)b(c−a)2=c2−bc+bc−a2(a−b)(b−c)(c−a)(2)
c(a−b)2=a2−ca+cb−c2(a−b)(b−c)(c−a)(3)c(a−b)2=a2−ca+cb−c2(a−b)(b−c)(c−a)(3)
Cộng theo vế (1);(2);(3) ta có ĐPCM
Lời giải:
Ta có:
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\Rightarrow \frac{a}{b-c}=\frac{-b}{c-a}+\frac{-c}{a-b}\)
\(\Leftrightarrow \frac{a}{b-c}=\frac{-b(a-b)-c(c-a)}{(a-b)(c-a)}=\frac{b^2+ca-c^2-ab}{(a-b)(c-a)}\)
\(\Rightarrow \frac{a}{(b-c)^2}=\frac{b^2+ca-c^2-ab}{(a-b)(b-c)(c-a)}\)
Hoàn toàn tương tự:
\(\frac{b}{(c-a)^2}=\frac{c^2+ab-a^2-bc}{(a-b)(b-c)(c-a)}\)
\(\frac{c}{(a-b)^2}=\frac{a^2+bc-b^2-ac}{(a-b)(b-c)(c-a)}\)
Cộng theo vế các đẳng thức vừa thu được ta có:
\(\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=\frac{b^2+ac-c^2-ab+c^2+ab-a^2-bc+a^2+bc-b^2-ac}{(a-b)(b-c)(c-a)}=0\)
Ta có đpcm.
Ta có:
\(a^2=\left(-b-c\right)^2\)
\(\Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự ta cũng có
\(\left\{{}\begin{matrix}b^2-c^2-a^2=2ca\\c^2-a^2-b^2=2ab\end{matrix}\right.\)
Thế vô ta được
\(A=\sqrt{\dfrac{3a^2}{bc}+\dfrac{3b^2}{ca}+\dfrac{3c^2}{ab}}\)
\(=\sqrt{\dfrac{3\left(a^3+b^3+c^3\right)}{abc}}\)
\(=\sqrt{3.\dfrac{\left(a^3+b^3+c^3-3abc\right)+3abC}{abc}}\)
\(=\sqrt{3.\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{abc}}\)
\(=\sqrt{3.3}=3\)
ĐPCM