1.

\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)

Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)

Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)

Từ đó ta được đpcm

2.

\(a,Sửa:a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^6-b^6\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^2-b^2+1\right)\\ =\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^2-b^2+1\right)\\ b,=\left(a^3+b^3\right)-1+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\\ =\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\\ =\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\)

\(c,=a^2b^2\left(b-a\right)+b^2c^2\left(c-a+a-b\right)-c^2a^2\left(c-a\right)\\ =-a^2b^2\left(a-b\right)+b^2c^2\left(a-b\right)+b^2c^2\left(c-a\right)-c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(b^2c^2-a^2b^2\right)+\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(c-a\right)\left(c+a\right)+c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(c-a\right)\left[b^2\left(c+a\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b^2c+ab^2-bc^2-ac^2\right)\\ =\left(a-b\right)\left(c-a\right)\left[bc\left(b-c\right)+a\left(b-c\right)\left(b+c\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(bc+ab+ac\right)\)

Ta có : \(\frac{2y+2z-x}{a}=\frac{2z+2x-y}{b}=\frac{2x+2y-z}{c}\)(sửa lại đề) (1) 

=> \(\frac{2y+2z-x}{a}=\frac{4b+4x-2y}{2b}=\frac{4x+4y-2z}{2c}\)

= \(\frac{4z+4x-2y+4x+4y-2z-2y-2z+x}{2b+2c-a}=\frac{9x}{2b+2c-a}\)(dãy tỉ số bằng nhau) (2)

Từ (1) => \(\frac{4y+4z-2x}{2a}=\frac{2z+2x-y}{b}=\frac{4x+4y-2z}{2c}\)

= \(\frac{4x+4y-2z+4y+4z-2x-2z-2x+y}{2c+2a-b}=\frac{9y}{2c+2a-b}\)(dãy tỉ số bằng nhau) (3)

Từ (1) có :  \(\frac{4y+4z-2x}{2a}=\frac{4z+4x-2y}{2b}=\frac{2x+2y-z}{c}=\frac{4y+4z-2x+4z+4x-2y-2x-2y+z}{2a+2b-c}\)\(=\frac{9z}{2a+2b-c}\)(dãy tỉ số bằng nhau) (4)

Từ (2) ; (3) ; (4) => điều phải chứng minh

1) \(\frac{x-y}{x+y}=\frac{z-x}{z+x}\)

\(\Leftrightarrow\left(x-y\right)\left(z+x\right)=\left(z-x\right)\left(x+y\right)\)

\(\Leftrightarrow z\left(x-y\right)+x\left(x-y\right)=x\left(z-x\right)+y\left(z-x\right)\)

\(\Leftrightarrow xz-zy+x^2-xy=xz-x^2+yz-xy\)

\(\Leftrightarrow-zy+x^2=-x^2+yz\)

\(\Leftrightarrow-2x^2=-2zy\)

\(\Leftrightarrow x^2=yz\)(đpcm)

\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)

\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)

\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)

b.

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)

\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)

1.

\(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c+2a+c}{2a+c}=\dfrac{a+b+c+2b}{2b}=\dfrac{a+b+c+b+c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}+1=\dfrac{a+b+c}{2b}+1=\dfrac{a+b+c}{b+c}+1\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}=\dfrac{a+b+c}{2b}=\dfrac{a+b+c}{b+c}\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

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

\(\Rightarrow\left\{{}\begin{matrix}2a+c=b+c\\2b=b+c\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a=b\\b=c\end{matrix}\right.\) \(\Rightarrow2a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+2a\right)\left(2a+2a\right)\left(2a+a\right)}{a.2a.2a}=9\)

Bài 2 đề sai

Ở phân thức thứ 2 không thể là \(\dfrac{y+3x-x}{x}\)

Bài 2:

\(P=\dfrac{x+3y}{y}\cdot\dfrac{y+3z}{z}\cdot\dfrac{z+3x}{x}=\dfrac{\left(x+3y\right)\left(y+3z\right)\left(z+3x\right)}{xyz}\)

Với \(x+y+z=0\)

\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}\\ \Leftrightarrow\dfrac{x+3y+x+y}{z}=\dfrac{y+3z+y+z}{x}=\dfrac{z+3x+x+z}{y}\\ \Leftrightarrow\dfrac{2\left(x+2y\right)}{z}=\dfrac{2\left(y+2z\right)}{x}=\dfrac{2\left(z+2x\right)}{y}\\ \Leftrightarrow\dfrac{2\left(y-z\right)}{z}=\dfrac{2\left(z-x\right)}{x}=\dfrac{2\left(x-y\right)}{y}\\ \Leftrightarrow\dfrac{2y-2z}{z}=\dfrac{2z-2x}{x}=\dfrac{2x-2y}{y}\\ \Leftrightarrow\dfrac{2y}{z}-2=\dfrac{2z}{x}-2=\dfrac{2x}{y}-2\\ \Leftrightarrow\dfrac{2y}{z}=\dfrac{2z}{x}=\dfrac{2x}{y}\\ \Leftrightarrow\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x}{y}\Leftrightarrow x=y=z=0\left(\text{trái với GT}\right)\)

Với \(x+y+z\ne0\)

\(\Leftrightarrow\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y-z=3z\\y+3z-x=3x\\z+3x-y=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=4z\\y+3z=4x\\z+3x=4y\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{4x\cdot4y\cdot4z}{xyz}=64\)