\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{1}{\left(a+b\right)^2}-\frac{2}{ab}}\)

\(=\sqrt{\left(\frac{a+b}{ab}\right)^2+\frac{1}{\left(a+b\right)^2}-\frac{2\left(a+b\right)}{ab}.\frac{1}{a+b}}\)

\(=\sqrt{\left(\frac{a+b}{ab}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)

làm nổi à bạn. 

1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )² = 0 \(\Rightarrow\)x² + y² + z² = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)

\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)

2. a + b + c = 1 \(\Rightarrow\)( a + b + c )² = 1 \(\Rightarrow\)a² + b² + c² + 2 ( ab + bc + ac ) = 1 \(\Rightarrow\)ab + bc + ac = 0

Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\)

\(\Rightarrow\)x = a ( x + y + z ) ; y = b ( x + y + z ) ; z = c ( x + y + z )

Ta có : xy + yz + xz = ab ( x + y + z )² + bc ( x + y + z )² + ac ( x + y + z )² = ( x + y + z )² ( ab + bc + ac ) = 0

3. sửa đề : 3x - y = 3z

Ta có : \(\hept{\begin{cases}3x-y=3z\\2x+y=7z\end{cases}\Rightarrow\hept{\begin{cases}\left(3x-y\right)+\left(2x+y\right)=3z+7z\\2x+y=7z\end{cases}\Rightarrow}\hept{\begin{cases}5x=10z\\y=7z-2x\end{cases}\Rightarrow}\hept{\begin{cases}x=2z\\y=3z\end{cases}}}\)

\(\Rightarrow\)\(S=\frac{x^2-2xy}{x^2+y^2}=\frac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\frac{4z^2-12z^2}{4z^2+9z^2}=\frac{-8z^2}{13z^2}=\frac{-8}{13}\)

Có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)(do a,b,c khác 0)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Suy ra: \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{3}{abc}\)(vì \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\))

Vậy...........

1. 

Ta có : \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)

\(\Rightarrow\frac{a.\left(2bz-3cy\right)}{a^2}=\frac{2b.\left(3cx-az\right)}{4b^2}=\frac{3c.\left(ay-2bx\right)}{9c^2}\)

\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)

Áp dụng tính chất của dãy tỉ số bằng hau ta có :

 \(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)

\(=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)

\(\Rightarrow\hept{\begin{cases}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz=3cy\\3cx=az\\ay=2bx\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{x}{3c}\left(đpcm\right)\)

Chúc bạn học tốt !!!

1. Sửa lại dòng cuối 

\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)

Bài 2:

Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)

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

\(\Rightarrow2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\)

\(\Rightarrow ab-bc=ac-ab\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)

Ta có : \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{1}{a}+\frac{1}{b}\right)\)

=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{a^3b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{a^2b^2}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{ab}\right)\)

=> \(P=\frac{1}{\left(a+b\right)^3}\left(\frac{a^3+b^3}{1}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{a^2+b^2}{1}\right)+\frac{6}{\left(a+b\right)^5}\left(\frac{a+b}{1}\right)\)

=> \(P=\frac{a^3+b^3}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2\right)}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)

=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a^2+b^2+2a\right)-6a}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}\)

=> \(P=\frac{\left(a+b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)^3}+\frac{3\left(a+b\right)^2}{\left(a+b\right)^4}+\frac{6\left(a+b\right)}{\left(a+b\right)^5}-\frac{6}{\left(a+b\right)^4}\)

=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}+\frac{6}{\left(a+b\right)^4}-\frac{6}{\left(a+b\right)^4}\)

=> \(P=\frac{a^2+ab+b^2}{\left(a+b\right)^2}+\frac{3}{\left(a+b\right)^2}=\frac{2a^2+4ab+2b^2}{\left(a+b\right)^2}-\frac{a^2+b^2}{\left(a+b\right)^2}\)

=> \(P=2-\frac{a^2+b^2}{\left(a+b\right)^2}=1+\frac{-2}{\left(a+b\right)^2}\)