2.

Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(1\right)\)

Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(dpcm\right)\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}=VP\)

Xảy ra khi \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

Ta có:

\(\dfrac{a.\left(x+z\right)}{abc}=\dfrac{b.\left(z+x\right)}{abc}=\dfrac{c.\left(x+y\right)}{abc}\)

\(\Rightarrow\dfrac{y+z}{bc}=\dfrac{x+z}{ac}=\dfrac{x+y}{ab}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{y+z}{bc}=\dfrac{x+z}{ac}=\dfrac{x+y}{ab}=\dfrac{z+x-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c.\left(a-b\right)}\left(1\right)\)

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

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

Từ \(\left(1\right),\left(2\right),\left(3\right)\) suy ra:

\(\dfrac{y-z}{a.\left(b-c\right)}=\dfrac{z-x}{b.\left(c-a\right)}=\dfrac{x-y}{c.\left(a-b\right)}\)

b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)

c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)

\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

Đề sai. Bạn xem lại.

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 \(x+y+z\ne0\)

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