Ta có : \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}\)\(=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{1}{2}\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

vãi cả loz sao lại sai ?

Ta có:

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{3+4+5}=\frac{x+y+z}{x+2y-z}.\left(\text{dãy tỉ số bằng nhau}\right)\)

\(\Rightarrow3+4+5=12=x+2y-z\)

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+2y-z}{3+8-5}=\frac{12}{6}=2\)

=> x/3 =2 =>x=6

=> x/4=2 => x=8

=> x/5 =2 => x=10

\(\left|x+1\right|+\left|x-2\right|=\left|x+1\right|+\left|2-x\right|\ge\left|x+1+2-x\right|=\left|3\right|=3\)

\(\Rightarrow\)\(3-\left(y-2\right)^2\ge3\)

\(\Leftrightarrow\)\(\left(y-2\right)^2\le0\)

\(\Leftrightarrow\)\(y-2=0\)

\(\Leftrightarrow\)\(y=2\)

\(PT\)\(\Leftrightarrow\)\(\left|x+1\right|+\left|x-2\right|\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x+1\right)\left(2-x\right)\ge0\)

TH1 : \(\hept{\begin{cases}x+1\ge0\\2-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le2\end{cases}\Leftrightarrow}-1\le x\le2}\)

TH2 : \(\hept{\begin{cases}x+1\le0\\2-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-1\\x\ge2\end{cases}}}\) ( loại ) 

Vậy \(-1\le x\le2\) và \(y=2\)

Chúc bạn học tốt ~ 

x-y=2;=0(1)-0(2)=2;0(1)=2+0;0(1)=2;0(2)=2-2;0(2)=0

vậy x=2;y=0

em mới lớp 6 à

Ta có : \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)

\(\Rightarrow\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\left(\frac{z}{3}\right)^2=\frac{x^2}{5^2}=\frac{y^2}{7^2}=\frac{z^2}{3^2}\)\(=\frac{x^2}{25}=\frac{y^2}{49}=\frac{z^2}{9}=\frac{x^2+y^2-z^2}{25+49-9}=\frac{585}{65}=9\)

\(\Rightarrow x=9.5=45\)

     \(y=9.7=63\)

     \(z=9.3=27\)