Ta có \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c=\dfrac{a+b+c}{2}\)

Suy ra

M = (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) + x²

    = x² - ax - bx + ab + x² - bx - cx + bc + x² - ax - cx + ac + x²

    = 4x² - 2ax - 2bx - 2cx + ab + bc + ac

    = (2x)² - 2x(a + b + c) + ab + bc + ac

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

    = ab + bc + ac

(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) + x²

= (x²-ax-bx+ab) + (x²-bx-cx+bc) + (x²-cx-ax+ac) + x²

= 4x² - 2ax - 2bx + ab + bc + ac

Thay a+b+c = 2x, ta được:

M = 4x² - 2x(a+b+c) + ab + bc + ac

M = 4x² - 2x.2x + ab + bc + ac

M = ab + bc + ac

Vậy => đcpcm

Bài 1 rút gọn bc tự làm :

\(B=\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)

\(B=\dfrac{3x^3-3y^2-4y^2+4y+y-1}{2y^3-2y^2+y^2-y+3y-3}\)

\(B=\dfrac{3y^2\left(y-1\right)-4y\left(y-1\right)+\left(y-1\right)}{2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)}\)

\(B=\dfrac{\left(3y^2-4y+1\right)\left(y-1\right)}{\left(2y^2+y-3\right)\left(y-1\right)}\)

\(B=\dfrac{3y^2-3y-y+1}{2y^2-2y+3y-3}=\dfrac{3y\left(y-1\right)-\left(y-1\right)}{2y\left(y-1\right)+3\left(y-1\right)}\)

\(B=\dfrac{\left(3y-1\right)\left(y-1\right)}{\left(3y+2\right)\left(y-1\right)}=\dfrac{3y-1}{3y+2}\)

Bài 2 )

a ) \(x+\dfrac{1}{x}=3\)

\(\Leftrightarrow x^2+2x\dfrac{1}{x}+\dfrac{1}{x^2}=9\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}=1\)

b ) \(\left(x+\dfrac{1}{x}\right)^3=27\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+\dfrac{3}{x}+3x=27\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(\dfrac{1}{x}+x\right)=27\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)

Bài 1:

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)

\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)

Bài 2:

a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)

b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)

1.

\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)

\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)

\(=\left(x^3-x^2+3x\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)

Hay đa thức trên có thể phân tích thành:

\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)

Dựa vào đó em tự tách cho phù hợp

2.

\(VT=a\left(\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+b\left(\dfrac{1}{a^2}+\dfrac{1}{c^2}\right)+c\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)

\(VT\ge\dfrac{2a}{bc}+\dfrac{2b}{ac}+\dfrac{2c}{ab}=2\dfrac{a^2+b^2+c^2}{abc}\)

\(VP=\dfrac{2\left(ab+bc+ca\right)}{abc}\)

\(\Rightarrow\dfrac{ab+bc+ca}{abc}\ge\dfrac{a^2+b^2+c^2}{abc}\)

\(\Rightarrow ab+bc+ca\ge a^2+b^2+c^2\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\le0\)

\(\Rightarrow a=b=c\)

3.

\(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\)

\(\Rightarrow\left(\dfrac{x^2-yz}{a}\right)^2=\left(\dfrac{y^2-xz}{b}\right)\left(\dfrac{z^2-xy}{c}\right)=\dfrac{\left(x^2-yz\right)^2-\left(y^2-xz\right)\left(z^2-xy\right)}{a^2-bc}\)

\(=\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}\)

Tương tự:

\(\left(\dfrac{y^2-xz}{b}\right)^2=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}\)

\(\left(\dfrac{z^2-xy}{c}\right)^2=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)

\(\Rightarrow\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)

\(\Rightarrow\dfrac{x}{a^2-bc}=\dfrac{y}{b^2-ac}=\dfrac{z}{c^2-ab}\Rightarrowđpcm\)

Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)

M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)

= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)

= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac

= 4x\(^2\)-2x(a+b+c)+ab+bc+ca

Thay 2x=a+b+c,ta được:

M= 4x\(^2\)-2x.2c+ab+bc+ca

M= 4x\(^2\)-4x\(^2\)+ab+bc+ca

M= ab+bc+ca

Câu 3: 

\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)

\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)

=>3x-2>0

=>x>2/3

Câu 1: 

a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)

\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)

\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)

\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)

\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)

\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)

\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)

\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)

\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)

b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)

TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)