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Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne̸0\) thì \(x=ak;y=bk;z=ck.\)
Do đó : \(\frac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)
\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{k^2\left(a^2+b^2+c^2\right)^2}{k^2\left(a^2+b^2+c^2\right)^2}=1.\)
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
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:
\(P=\frac{\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{a^2k^2+b^2k^2+c^2k^2}{\left[k\left(a^2+b^2+c^2\right)\right]^2}=\frac{k^2\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\frac{1}{a^2+b^2+c^2}\)
theo đề bài: \(ax+by+cz=0\)=> \(\left(ax+by+cz\right)^2=0\)
=> \(a^2x^2+b^2y^2+c^2z^2+2\left(axby+bycz+axcz\right)=0\left(1\right)\)
ta lại có tử số =\(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)
=\(bcy^2+bcz^2+caz^2+acx^2+abx^2+aby^2-2\left(abxy+acxz+bcyz\right)\)(2)
từ (1)(2)=>
Tử số=\(ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)+a^2x^2+b^2y^2+c^2z^2\)
=\(\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)\)
vậy A=a+b+c
Giải:
Chú ý sử dụng hằng đẳng thức :
(m+n+p)2 = m2 + n2 + p2 + 2mn + 2mp + 2np
Áp dụng hằng đẳng thức trên, ta có:
ax+by+cz = 0 ⇒ (ax+by+cz)2 = 0
⇒a2x2+b2y2+c2z2+2ax.by+2ax.cz+2by.cz=0
⇒a2x2+b2y2+c2z2= − (2abxy+2aczx+2bcyz)
Ta lại có:
bc.(y−z)2+ac.(x−z)2+ab.(x−y)2
=bc(y2−2yz+z2)+ac(x2−2xz+z2)+ab(x2−2xy+y2)
=bcy2+bcz2−2bcyz+acx2+acz2−2acxz+abx2+aby2−2abxy
=(bcy2+bcz2+acx2+acz2+abx2+aby2)−(2abxy+2aczx+2bcyz)
=bcy2+bcz2+acx2+acz2+abx2+aby2+a2x2+b2y2+c2z2
=x2(ac+ab+a2)+y2(bc+ab+b2)+z2(bc+ac+c2)
=ax2(a+b+c)+by2(a+b+c)+cz2(a+b+c)
=(a+b+c)(a.x2+b.y2+c.z2)
Vậy:
A=a.x2+b.y2+c.z2bc.(y−z)2+ac.(x−z)2+ab.(x−y)2=ax2+by2+cz2(a+b+c)(a.x2+b.y2+c.z2)
A=1a+b+cA=1a+b+c
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 )
Mà \(a.x+by+cz=0\)
\(\Rightarrow\left(a.x+by+cz\right)^2=0^2\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2\left(axby+axcz+bycz\right)=0\)( 2 )
\(\left(1\right)\left(2\right)\Rightarrow B=B+0\)
\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)+a^2x^2+b^2y^2+c^2z^2+2\left(axby+axcz+bycz\right)\)
\(=a.x^2\left(b+c\right)+b.y^2\left(a+c\right)+c.z^2\left(a+b\right)+a^2x^2+b^2y^2+z^2c^2\)
\(=a.x^2\left(a+b+c\right)+b.y^2\left(a+b+c\right)+cz^2\left(a+b+c\right)\)
\(=\left(a.x^2+by^2+cz^2\right)\left(a+b+c\right)\)
\(\Rightarrow A=\frac{B}{ax^2+by^2+cz^2}=a+b+c\)
Vậy ...
Ta có : \(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\Rightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2czax=0\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2abxy-2bycz-2czax\)
Xét tử số : \(bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(z^2-2xz+x^2\right)+ab\left(x^2-2xy+y^2\right)\)\(=bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(bcyz+abxy+acxz\right)\)
\(=bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)
\(=c\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+a\left(ax^2+by^2+cz^2\right)\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
\(\Rightarrow A=\frac{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}=\frac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)