Đặt: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow x=ak;y=bk;z=ck\)

Thay các kết quả trên vào P ta có:

\(P=\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2}{\left(a.ak+b.bk+c.ck\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{[k\left(a^2+b^2+c^2\right)]^2}=\dfrac{k^2\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

1) Đặt \(B=x^2+y^2+z^2\)

\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)

Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)

\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)

Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)

\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)

\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)

\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)

Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

Giả sử điều cần c/m là đúng . Khi đó , ta có :

\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

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

\(\Leftrightarrow x^2a^2+y^2a^2+z^2a^2+x^2b^2+y^2b^2+z^2b^2+x^2c^2+y^2c^2+z^2c^2\)

\(=x^2a^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2=2axby+2bycz+2axcz\)

\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2-2axby-2bycz-2axcz=0\) \(\Leftrightarrow\left(y^2a^2-2axby+b^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)+\left(x^2c^2-2axcz+a^2z^2\right)=0\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2=0\left(1\right)\)

Do \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\\\left(cx-az\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2\ge0\left(2\right)\)

Từ ( 1 ) ; ( 2 ) \(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\bz-cy=0\\cx-az=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\bz=cy\\cx=az\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\) \(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

Điều này đúng với GT đề bài cho

\(\Rightarrow\) Điều cần c/m là đúng

\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\left(đpcm\right)\)

hơi dài bạn ạ bđt trên đúng theo bunhia vì dấu "=" đúng với điều kiện rồi

Đặt \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{1}{k}\Rightarrow x=ak;y=bk;y=ck\)

\(\Rightarrow\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{a^2k^2+b^2k^2+c^2k^2}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{k^2\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\frac{1}{a^2+b^2+c^2}\)

Mạo phép sửa đề!CMR: \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{3}{a^2+b^2+c^2}\)

Ta có: \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) 

\(\Rightarrow\frac{x^2}{ax}=\frac{y^2}{by}=\frac{z^2}{cz}=\frac{x^2+y^2+z^2}{ax+by+cz}\)  (t/c dãy tỉ số bằng nhau)

\(\Rightarrow\frac{x^2}{\left(ax\right)^2}=\frac{y^2}{\left(by\right)^2}=\frac{z^2}{\left(cz\right)^2}=\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}\) (1)

Lại có: \(\frac{x^2}{\left(ax\right)^2}=\frac{y^2}{\left(by\right)^2}=\frac{z^2}{\left(cz\right)^2}=\) \(\frac{x^2}{a^2x^2}=\frac{y^2}{b^2y^2}=\frac{z^2}{c^2z^2}=\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{3}{a^2+b^2+c^2}\)

\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)

\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)

Dấu "=" xảy ra khi \(x=y\)

\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)

\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)

\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)

\(B_{min}=1\) khi \(x=y\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)

Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

a, \(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)=\(\dfrac{x^2+2xy+y^2}{4\left(x+y\right)}\)   = \(\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\)  =\(\dfrac{x+y}{4}\) 

a. \(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)

\(=\dfrac{x^2+2xy+y^2}{4\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\)

\(=\dfrac{x+y}{4}\)

b. \(\dfrac{x+5}{2x-2}-\dfrac{4}{x^2-1}:\dfrac{2}{x+1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{4}{\left(x+1\right)\left(x-1\right)}:\dfrac{2}{x+1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{2}{x-1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{4}{2\left(x-1\right)}\)

\(=\dfrac{x+1}{2\left(x-1\right)}\)