Đặt B = \(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)\) (1)

Từ \(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)

=>\(a^2x^2+b^2y^2+c^2z^2+2\left(bcyz+acxz+abxy\right)=0\)

=>\(a^2x^2+b^2y^2+c^2z^2=-2\left(bcyz+acxz+abxy\right)\) (2)

Thay (2) vào (1) ta được:

\(B=ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)+a^2x^2+b^2y^2+c^2z^2\)

\(=ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)\)

\(=\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)\)

Vậy \(A=\frac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{c}{z}=k\ne0\) thì \(x=ak;y=bk;z=ck.\)

Do đó : \(\frac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)

\(=\frac{k^2\left(a^2+b^2+c^2\right)^2}{k^2\left(a^2+b^2+c^2\right)^2}=1.\)

ai tích cho mình đi,mình tích lại cho

\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne̸0\) thì \(x=ak;y=bk;z=ck.\)

Do đó : \(\frac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)

\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{k^2\left(a^2+b^2+c^2\right)^2}{k^2\left(a^2+b^2+c^2\right)^2}=1.\)

Trường hợp không nghĩ ra cách nào hay và gọn để làm, ta đặt

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

\(\Rightarrow x=am;\text{ }y=bm;\text{ }z=cm\)

\(P=\frac{a^2m^2+b^2m^2+c^2m^2}{\left(a^2m+b^2m+c^2m\right)^2}=\frac{\left(a^2+b^2+c^2\right)m^2}{\left(a^2+b^2+c^2\right)^2.m^2}=\frac{a^2+b^2+c^2}{\left(a^2+b^2+c^2\right)^2}=\frac{1}{a^2+b^2+c^2}\)

Phân tích mẫu :

\(M=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

Khai triển các bình phương và gom các nhân tử chung :

\(M=\left(ab+ac\right)x^2+\left(ab+bc\right)y^2+\left(bc+ac\right)z^2-2abxy-2bcxy-2acxy\)

\(=\left[\left(ab+ac\right)x^2+a^2x^2+\left(ab+bc\right)y^2+b^2y^2+\left(bc+ac\right)z^2+c^2z^2\right]-\)\(\left(a^2x^2+b^2y^2+c^2z^2+2ab+2aczx+2bcyz\right)\)

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)-\left(ax+by+cz\right)^2\)

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\) ( vì \(ax+by+cz=0\) )

Kết quả :  \(M=\frac{1}{a+b+c},a+b+c\ne0\)

Đặt biểu thức trên là A
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne0\)

\(\Rightarrow x=ak,y=bk,z=ck\)

Nên \(A=\frac{\text{[}\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2\text{]}.\left(a^2+b^2+c^2\right)}{\left(a.ak+b.bk+c.bk\right)^2}\)

\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right).\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(=\frac{k^2\left(a^2+b^2+c^2\right).\left(a^2+b^2+c^2\right)}{\text{[}k\left(a^2+b^2+c^2\right)\text{]}^2}\)

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

\(=1\)

Vậy A=1

à quên sửa dòng trên chỗ A=1 cái chỗ mẫu là \(k^2.\left(a^2+b^2+c^2\right)^2\)nhen :v