Ta có:

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

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

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

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

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

Vì \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\) nên

\(\Rightarrow\hept{\begin{cases}\left(ay-bx\right)^2=0\\\left(az-cx\right)^2=0\\\left(bz-cy\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}\Leftrightarrow}\hept{\begin{cases}ay=bx\\az=cx\\bz=cy\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{a}{x}=\frac{c}{z}\\\frac{b}{y}=\frac{c}{z}\end{cases}}\)( Theo tính chất của tỉ lệ thức : tích trung tỉ bằng tích ngoại tỉ)

\(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)( Với x, y, z khác 0)

Vậy \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)Với x, y, z khác 0

Ta có:

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

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

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

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

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

Vì \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\) nên

\(\Rightarrow\hept{\begin{cases}\left(ay-bx\right)^2=0\\\left(az-cx\right)^2=0\\\left(bz-cy\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}\Leftrightarrow}\hept{\begin{cases}ay=bx\\az=cx\\bz=cy\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{a}{x}=\frac{c}{z}\\\frac{b}{y}=\frac{c}{z}\end{cases}}\)( Theo tính chất của tỉ lệ thức : tích trung tỉ bằng tích ngoại tỉ)

\(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)( Với x, y, z khác 0)

Vậy \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)Với x, y, z khác 0

Ta có:

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

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

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

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

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

Vì \(\left(ay-bx\right)^2\ge0;\left(az-cx\right)^2\ge0;\left(bz-cy\right)^2\ge0\) nên

\(\Rightarrow\hept{\begin{cases}\left(ay-bx\right)^2=0\\\left(az-cx\right)^2=0\\\left(bz-cy\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}\Leftrightarrow}\hept{\begin{cases}ay=bx\\az=cx\\bz=cy\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{a}{x}=\frac{c}{z}\\\frac{b}{y}=\frac{c}{z}\end{cases}}\)( Theo tính chất của tỉ lệ thức : tích trung tỉ bằng tích ngoại tỉ)

\(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)( Với x, y, z khác 0)

Vậy \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)Với x, y, z khác 0

Do x + y + z = 0 nên

x = - (y + z) ; y = - (x + z) ; z = - (x + y)

=> x² = (y + z)² ; y² = (x + z)² ; z² = (x + y)²

=> ax² + by² + cz² = a(y² + 2yz + z²) + b(x² + 2xz + z²) + c(x² + 2xy + y²) = x²(b + c) + y²(a + c) + z²(a + b) + 2(ayz + bxz + cxy)              (1) 

Thay a = - (b + c) ; b = - (a + c) ; c = - (a + b) (Do a + b + c = 0 ) và ayz+bxz+cxy=0 (do a/x+b/y+c/z=0) vào (1) ta được ax² + by² + cz² = - (ax² + by² + cz²)

=> ax² + by² + cz² = 0

to ko hieu noi

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

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

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz\)\(=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)

\(\Rightarrow b^2x^2-2abxy+a^2y^2+b^2z^2-2bcyz+c^2y^2+a^2z^2-2acxz+c^2x^2=0\)

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

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

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

\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2\left(abxy+bcyz+cazx\right)=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)\(\Leftrightarrow a^2y^2-2ay\cdot bx+b^2x^2+b^2z^2-2bz\cdot cy+c^2y^2+a^2z^2-2az\cdot cx+c^2x^2=0\)

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

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

Ta có \(ax^3=by^3=cz^3\Leftrightarrow\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=ax^2+by^2+cz^2\Leftrightarrow\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}=\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}+\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}+\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)Vậy \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)

Phá ngoặc hết ra rồi phân tích thành tổng 3 bình phương.

Câu hỏi của nguyễn ngọc minh - Toán lớp 8 - Học toán với OnlineMath