TA có: \(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b} = 0 \quad (1)\)

=>\(\frac{a}{b-c}+\frac{b}{c-a}=\frac{-c}{a-b}\)

=>\(\frac{a(c-a) + b(b-c)}{(b-c)(c-a)} = -\frac{c}{a-b}\)

=>\(\frac{ac - a^2 + b^2 - bc}{(b-c)(c-a)}=-\frac{c}{a-b}\)

=>\(\frac{(b^2 - a^2) - c(b - a)}{(b-c)(c-a)}=-\frac{c}{a-b}\)

=>\(\frac{(b - a)(a + b - c)}{(b-c)(c-a)}=-\frac{c}{a-b}\)

=>\(\frac{-(a - b)(a + b - c)}{(b-c)(c-a) \cdot (a - b)} = -\frac{c}{(a - b)^2}\)

=>\(\frac{-(a + b - c)}{(b-c)(c-a)}=-\frac{c}{(a - b)^2}\)

=>\(\frac{c}{(a-b)^2}=\frac{a + b - c}{(b-c)(c-a)}\quad\)(2)

Chứng minh tương tự, ta cũng có:

\(\frac{a}{(b-c)^2}=\frac{b + c - a}{(c-a)(a-b)}\quad\left(3\right)\)

\(\frac{b}{(c-a)^2}=\frac{c + a - b}{(a-b)(b-c)}\quad\left(4\right)\)

Từ (2),(3),(4) suy ra \(S=\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=\frac{b + c - a}{(c-a)(a-b)}+\frac{c + a - b}{(a-b)(b-c)}+\frac{a + b - c}{(b-c)(c-a)}\)

\(=\frac{(b+c-a)(b-c) + (c+a-b)(c-a) + (a+b-c)(a-b)}{(a-b)(b-c)(c-a)}\)

\(=\frac{b^2-c^2-ab+ac+c^2-a^2-bc+ab+a^2-b^2-ac+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

=0

Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...

Ta có: \(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1\)

=>\(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)

=>\(a \cdot \frac{a}{b+c} + (b+c) \cdot \frac{a}{b+c} + b \cdot \frac{b}{c+a} + (c+a) \cdot \frac{b}{c+a} + c \cdot \frac{c}{a+b} + (a+b) \cdot \frac{c}{a+b} = a + b + c\)

=>\(\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)

=>\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

Vì a>2=>a tru 2>0 (1)

b>2=>b tru 2>0 (2)

Ta có: 2ab tru 2.(a+b)=ab+ab tru 2a tru 2b=(ab tru 2a)+(ab tru 2b)=a(b tru 2)+b(a tru 2) (3)

Vì a,b khac 0 (4)

Từ (1),(2),(3),(4)=>2ab tru 2(a+b)>0

=>2ab>2(a+b)=>ab>a+b(đpcm)

Bai hoi kho nhin vi ban phim cua mk ko an dc dau tru. Ban thog cam nha :-)

\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)

\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

Cho \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với ( với a, b, c, d khác 0, và c \(\ne\pm d\) ). Chứng minh rằng hoặc \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\) ?

a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)

\(=a-1\)

b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)

\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)

c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)

\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)

                      Bài làm :

1) Khi x=9 ; giá trị của A là :

\(A=\frac{\sqrt{9}}{\sqrt{9}+2}=\frac{3}{3+2}=\frac{3}{5}\)

2) Ta có :

\(B=...\)

\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)

\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)

3) Ta có :

\(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+2}\div\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)

Xét :

\(\frac{A}{B}+1=\frac{4}{\sqrt{x+2}}>0\Rightarrow\frac{A}{B}>-1\)

=> Điều phải chứng minh

1, thay x=9(TMĐKXĐ) vào A ta đk:

A=\(\dfrac{\sqrt{9}}{\sqrt{9}-2}=3\)

vậy khi x=9 thì A =3

2,với x>0,x≠4 ta đk:

B=\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

3,\(\dfrac{A}{B}>-1\) (x>0,x≠4)

⇒\(\dfrac{\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}>-1\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}>-1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>-1\)

⇒\(\sqrt{x}-2>-1\) (vì \(\sqrt{x}+2>0\))

⇔\(\sqrt{x}>1\)⇔x=1 (TM)

vậy x=1 thì \(\dfrac{A}{B}>-1\) với x>0 và x≠4

Đặt x=a-b; y=b-c; z=c-a

=>x+y+z=0

a+b+c=0

=>a+b=-c; b+c=-a; a+c=-b

Đặt \(X = \frac{a-b}{c} + \frac{b-c}{a} + \frac{c-a}{b}\)

\(=\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{abc}\)

Đặt \(Y = \frac{c}{a-b} + \frac{a}{b-c} + \frac{b}{c-a}\)

\(=\frac{c}{x}+\frac{a}{y}+\frac{b}{z}=\frac{cyz+axz+bxy}{xyz}\)

Đặt T=cyz+axz+bxy

\(=cyz+(-b-c)xz+bxy=cyz-bxz-cxz+bxy\)

\(=cz(y-x)+bx(y-z)=-3abc\)

Do đó: \(Y = \frac{-3abc}{(a-b)(b-c)(c-a)}\)

A=X*Y

\(=\frac{-\left(a-b\right)\cdot\left(b-c\right)\left(c-a\right)}{abc}\cdot\frac{-3abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

=3

\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)