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Áp dụng tính chất dãy tỉ số bằng nhau ta có: \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=k\)
\(\Rightarrow\hept{\begin{cases}a=kx;b=ky;c=kz\Rightarrow a^2=k^2x^2;b^2=k^2y^2;c^2=k^2z^2\\a+b+c=k\left(x+y+z\right)\end{cases}}\)
Có: \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{x^2+y^2+z^2}{\left(kx^2+ky^2+kz^2\right)^2}=\frac{x^2+y^2+z^2}{k^2\left(x^2+y^2+z^2\right)^2}=\frac{1}{k^2\left(x^2+y^2+z^2\right)}\)
\(=\frac{1}{k^2x^2+k^2y^2+k^2z^2}=\frac{1}{a^2+b^2+c^2}\)(đpcm)
\(\text{Áp dụng BĐT Bunhia... cho 2 bộ số (a;b;c) và (x;y;z), ta có: }\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\ge\left(ax+by+cz\right)^2\)
\(\text{Dấu = xảy ra }\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\text{(đpcm)}\)
Chả biết có đúng không '-'
Sửa lại đề:\(\left(ax+by+cz\right)\rightarrow\left(ax+by+cz\right)^2\)
Ta có:\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2aybx-2bzcy-2azcx=0\)
\(\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
Vì\(\left(ay-bx\right)^2\ge0\)
\(\left(bz-cy\right)^2\ge0\)
\(\left(az-cx\right)^2\ge0\)
Suy ra:\(\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2\ge0\)
Mà\(\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\bz-cy=0\\az-cx=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}ay=bx\\bz=cy\\az=cx\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}}\)\(\left(x,y,z\ne0\right)\)
\(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\left(đpcm\right)\)
Vậy...
Linz
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\left(ax+by+cz\right)^2=\left(x^2+y^2+z^2\right)\left(a^2+b^2\right)+c^2\)
Khai triển và rút gọn ta được:
\(2abxy+2acxz+2bcyz=a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2\)
Gom các hạng tử lai thành hằng đẳng thức:
\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(cy-bz\right)^2=0\)
Mà bình phương của các số thì \(\ge\)0 nó chỉ bằng 0 khi và chỉ khi từng hạng tử =0
nên: \(\hept{\begin{cases}ay-bx=0\\az-cx=0\\cy-bz=0\end{cases}\Rightarrow\hept{\begin{cases}ay=bx\rightarrow\frac{a}{x}=\frac{b}{y}\\az=cx\rightarrow\frac{a}{x}=\frac{c}{z}\\cy=bz\rightarrow\frac{c}{z}=\frac{b}{y}\end{cases}}}\)
=>Đpcm
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Rightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz\)\(=a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(\Rightarrow b^2x^2-2abxy+a^2y^2+b^2z^2-2bcyz+c^2y^2+a^2z^2-2acxz+c^2x^2=0\)
\(\Rightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}bx-ay=0\\bz-cy=0\\az-cx=0\end{cases}\Rightarrow\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{b}{y}=\frac{a}{x}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}\Rightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}}\)
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2\left(abxy+bcyz+cazx\right)=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)\(\Leftrightarrow a^2y^2-2ay\cdot bx+b^2x^2+b^2z^2-2bz\cdot cy+c^2y^2+a^2z^2-2az\cdot cx+c^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
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) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)
\(\Leftrightarrow b^2x^2-2abxy+a^2y^2=0\)
\(\Leftrightarrow\left(bx\right)^2-2\cdot bx\cdot ay+\left(ay\right)^2=0\)
\(\Leftrightarrow\left(bx-ay\right)^2=0\Rightarrow bx=ay\Rightarrow\left(\frac{a}{x}=\frac{b}{y}\right)\)
b) \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2acxz\)
\(\Leftrightarrow b^2x^2-2bxay+a^2y^2+b^2z^2-2bzcy+c^2y^2+a^2z^2-2azcx+c^2x^2=0\)
\(\Leftrightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
\(\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}}\Rightarrow\left(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\right)}\)
c) \(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+b^2+2ab=2a^2+2b^2\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\)
a, Tương đương : \(a^2x^2+a^2y^2+b^2x^2+b^2y^2\) = \(a^2x^2+2axby+b^2y^2\)
\(a^2y^2-2axby+b^2x^2=0\)
\(\left(ay-bx\right)^2\) = 0
\(ay-bx=0\)
\(ay=bx\)
\(\frac{a}{x}=\frac{b}{y}\) dpcm
Câu b, c làm tương tự câu a
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
Mà \(\left(ax+by+cz\right)^2=a^2x^2+b^2y^2+c^2x^2+2abxy+2acxz+2bcyz\)
Nên \(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2acxz+2bcyz\)
\(\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2abxy-2acxz-2bcyz=0\)
\(\Leftrightarrow\left(a^2y^2-2abxy+b^2x^2\right)+\left(a^2z^2-2acxz+c^2x^2\right)+\left(b^2z^2-2bcyz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}\Rightarrow\hept{\begin{cases}ay=bx\\az=cx\\bz=cy\end{cases}\Rightarrow}\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{a}{x}=\frac{c}{z}\\\frac{b}{y}=\frac{c}{z}\end{cases}}}\)
\(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\) (đpcm)