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

=> x = ak, y = bk, z = ck

Thay x = ak, y = bk, z = ck vào P, ta có:

\(P=\frac{\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{a^2k^2+b^2k^2+c^2k^2}{\left[k\left(a^2+b^2+c^2\right)\right]^2}=\frac{k^2\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\frac{1}{a^2+b^2+c^2}\)

Với điều kiện như đề bài

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

Tướng tự: 

\(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c-b}{b+a}+\frac{b-a}{b+c}\)

\(\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\)

Em nhớ làm tiếp nhé!

làm tiếp kiểu gì ạ 

\(\frac{x^4-4x^2+3}{x^4-6x^2-7}\)

\(=\frac{x^4-x^2-3x^2+3}{x^4-x^2+7x^2-7}\)

\(=\frac{x^2.\left(x^2-1\right)-3.\left(x^2-1\right)}{x^2.\left(x^2-1\right)+7.\left(x^2-1\right)}\)

\(=\frac{\left(x^2-3\right).\left(x^2-1\right)}{\left(x^2+7\right).\left(x^2-1\right)}=\frac{x^2-3}{x^2+7}\)

\(x^2-6x+5+\left(x-5\right)^2\)

\(=x^2-6x+5+x^2-10x+25\)

\(=2x^2-6x-10x+30\)

\(=x.\left(2x-6\right)-5.\left(2x-6\right)\)

\(=\left(x-5\right).\left(2x-6\right)\)

\(x^3=x\)

=> \(x^3-x=0\)

=> \(x\left(x^2-1\right)=0\)

=> \(\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow[\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)

x^3 =x

x*x*x =x

Suy ra :x ={1;-1;0}