Áp dụng BĐT \(x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\) ta có:

\(VT_{Pt\left(2\right)}=x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}=0,75=VP_{Pt\left(2\right)}\)

Xảy ra khi \(x=y=z=\dfrac{1}{2}\)

\(x^2+y^2+z^2-x-y-z+0,75=0\)

\(\Leftrightarrow x^2+y^2+z^2-x-y-z+\frac{3}{4}=0\)

\(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)+\left(z^2-z+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2=0\)

Dễ thấy: \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{1}{2}\right)^2\ge0\)

Xảy ra khi \(\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\\z-\frac{1}{2}=0\end{cases}}\)\(\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\\z-\frac{1}{2}=0\end{cases}}\)

bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

x=25;y=75;z=125

a) \(\frac{x}{2}=\frac{y}{3}\)và \(\frac{y}{5}=\frac{z}{7}\)và \(x+y+z=92\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chất dãy tỉ số = nhau

ta có 

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)

Suy ra \(\frac{x}{10}=2\Rightarrow x=2.10=20\)

\(\frac{y}{15}=2\Rightarrow y=2.15=30\)

\(\frac{z}{21}=2\Rightarrow z=2.21=42\)

Vậy \(x=20;y=30;z=42\)

bài 2 :

ta có x:y:z=3:5:(-2)

=>x/3=y/5=z/-2

=>5x/15=y/5=3z/-6

áp dụng tc dãy ... ta có :

5x/15=y/5=3z/-6=5x-y+3z/15-5+(-6)=-16/4=-4

=>x/3=-=>x=-12

=>y/5=-4=>y=-20

=>z/-2=-4=>z=8