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

\(\Rightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)

\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\left(1\right)\)

*)Xét \(x+y+z\ne0\left(2\right)\). Từ (1) và (2)

\(\Rightarrow x=y=z\). Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=2\cdot2\cdot2=8\)

*)Xét \(x+y+z=0\)\(\Rightarrow\left\{\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

Khi đó \(B=\frac{x+y}{y}\cdot\frac{y+z}{z}\cdot\frac{x+z}{x}=\frac{-z}{y}\cdot\frac{-x}{z}\cdot\frac{-y}{x}=-1\)

a)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có

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

\(\Rightarrow\left\{\begin{matrix}\frac{y+z-x}{x}=1\\\frac{z+x-y}{y}=1\\\frac{x+y-z}{z}=1\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\) (1)

Ta có \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)

\(\Rightarrow B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)

Thế (1) vào biểu thức B

\(\Rightarrow B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}\)

\(\Rightarrow B=2.2.2=8\)

Vậy biểu thức \(B=8\)

mik đag cần gấp các bn giải nhanh dùm mik nha

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)

Ta có: \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\frac{ak\cdot bk\cdot ck\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\cdot\left(ak+bk\right)\cdot\left(bk+ck\right)\cdot\left(ck+ak\right)}\)

\(=\frac{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3\cdot abc\cdot\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)

Vậy: H=1

đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)

theo giả thiết ta có \(H=\frac{xyz\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

thay \(H=\frac{ak.bk.ck\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(ak+bk\right)\left(bk+ck\right)\left(ck+ak\right)}\)

\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left[k\left(a+b\right)\right]\left[k\left(b+c\right)\right]\left[k\left(c+a\right)\right]}\)

\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc.k\left(a+b\right).k\left(b+c\right).k\left(c+a\right)}\)

\(\Leftrightarrow H=\frac{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}{k^3abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}=1\)

Vậy H = 1

Hay quớ ak! Mơn m nhìu nha ný! <3 <3 <3 (not thả thính =))))

chỉ thả tai thui

1a.

\(2P=1-\frac{bc}{2a^2+bc}+1-\frac{ca}{2b^2+ca}+1-\frac{ab}{2c^2+ab}\)

\(\Rightarrow2P=3-\left(\frac{bc}{2a^2+bc}+\frac{ca}{2b^2+ca}+\frac{ab}{2c^2+ab}\right)\)

\(\Rightarrow2P=3-\left(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{c^2a^2}{2b^2ca+c^2a^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\right)\)

\(\Rightarrow2P\le3-\frac{\left(ab+bc+ca\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=3-1=2\)

\(\Rightarrow P\le1\)

\(P_{max}=1\) khi \(a=b=c\)

1b.

\(Q=\frac{a^2}{5a^2+b^2+c^2+2bc}+\frac{b^2}{5b^2+a^2+c^2+2ca}+\frac{c^2}{5c^2+a^2+b^2+2ab}\)

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

\(\Rightarrow Q\le\frac{1}{9}\left(\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)

\(\Rightarrow Q\le\frac{1}{9}\left(1+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)

Theo kết quả câu a ta có:

\(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\le1\)

\(\Rightarrow Q\le\frac{1}{9}\left(1+2\right)=\frac{1}{3}\)

\(Q_{max}=\frac{1}{3}\) khi \(a=b=c\)

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

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

Áp dụng bđt AM-GM ta có

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

\(\ge27.\frac{a^2+b^2+c^2+ab+bc+ca}{8\left(a+b+c\right)^3}\)\(=\frac{a^2+b^2+c^2+ab+bc+ca}{8}\)

\(=\frac{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}{8}\)\(\ge\frac{9-\frac{\left(a+b+c\right)^2}{3}}{8}=\frac{9-3}{8}=\frac{3}{4}\)

Dấu "=" xảy ra khi a=b=c=1

a/ Một cách đơn giản hơn:

\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

\(P=\frac{x-\frac{1}{2}+y-\frac{1}{2}}{y^2}+\frac{y-\frac{1}{2}+z-\frac{1}{2}}{z^2}+\frac{z-\frac{1}{2}+x-\frac{1}{2}}{x^2}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(P=\left(x-\frac{1}{2}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\left(y-\frac{1}{2}\right)\left(\frac{1}{y^2}+\frac{1}{z^2}\right)+\left(z-\frac{1}{2}\right)\left(\frac{1}{x^2}+\frac{1}{z^2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(P\ge\frac{2}{xy}\left(x-\frac{1}{2}\right)+\frac{2}{yz}\left(y-\frac{1}{2}\right)+\frac{2}{zx}\left(z-\frac{1}{2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(P\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\)

\(P\ge\sqrt{3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}-1=\sqrt{3}-1\)

\(P_{min}=\sqrt{3}-1\) khi \(x=y=z=\sqrt{3}\)