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\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Rightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz\)\(=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Rightarrow b^2x^2-2abxy+a^2y^2+b^2z^2-2bcyz+c^2y^2+a^2z^2-2acxz+c^2x^2=0\)
\(\Rightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}bx-ay=0\\bz-cy=0\\az-cx=0\end{cases}\Rightarrow\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{b}{y}=\frac{a}{x}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}\Rightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}}\)
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2\left(abxy+bcyz+cazx\right)=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)\(\Leftrightarrow a^2y^2-2ay\cdot bx+b^2x^2+b^2z^2-2bz\cdot cy+c^2y^2+a^2z^2-2az\cdot cx+c^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
mà \(\left(ay-bx\right)^2;\left(bz-cy\right)^2;\left(az-cx\right)^2\ge0\)nên \(\left(ay-bx\right)^2=\left(bz-cy\right)^2=\left(az-cx\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}ay=bx\\bz=cy\\az=cx\end{cases}\Leftrightarrow\frac{a}{x}}=\frac{b}{y}=\frac{c}{z}\left(x,y,z\ne0\right)\)(ĐPCM)
Bạn ko hiểu chỗ nào cứ hỏi lại mình nhé
\(a+\frac{1}{b}=1\)\(\Leftrightarrow\left(a+\frac{1}{b}\right)^2=1\)\(\Leftrightarrow a^2+\frac{1}{b^2}+\frac{2a}{b}=1\)\(\Leftrightarrow\frac{a}{b}=-1\)
\(a^2+\frac{1}{b^2}=3\)\(\Leftrightarrow\left(a^2+\frac{1}{b^2}\right)^2=9\)\(\Leftrightarrow a^4+\frac{1}{b^4}+\frac{2.a^2}{b^2}=9\)\(\Leftrightarrow a^4+\frac{1}{b^4}=7\)
\(N=\frac{a^4b^4+a^2b^2+1}{b^4}=a^4+\frac{a^2}{b^2}+\frac{1}{b^4}\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
\(\frac{\Leftrightarrow a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\) Nhân hai vế với \(\frac{1}{b-c}\)
Tương tự ta có:\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right);\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng (1),(2),(3) ta được đpcm
