Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

VP=\(A^2X^2+B^2Y^2+C^2Z^2+A^2Y^2+B^2X^2+A^2Z^2+C^2X^2+B^2Z^2+C^2Y^2\)

=\(A^2\left(X^2+Y^2+Z^2\right)+B^2\left(X^2+Y^2+Z^2\right)+C^2\left(X^2+Y^2+Z^2\right)\)

=\(\left(X^2+Y^2+Z^2\right)\left(A^2+B^2+C^2\right)\)

\(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}\)

\(\Rightarrow\frac{acy-bcx}{c^2}=\frac{bcx-abz}{b^2}=\frac{abz-acy}{a^2}=\frac{0}{a^2+b^2+c^2}=0\)

\(\Rightarrow\hept{\begin{cases}ay-bx=0\\cx-az=0\\bz-cy=0\end{cases}}\)

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

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

\(+c^2y^2=0\)

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

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

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

Bình phương ba vế suy ra \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

Sau đó chứng minh tương tự bunhiacopxki

a) Sửa đề: \(\left(ax+by+cx\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2\)
= a²x² + b²y² + c²x² + 2axby + 2bycz + 2axcz + b²x² - 2bxay + a²y² + c²y² - 2cybz + b²z² + a²z² - 2azcx + c²x²
= a²x² + b²y² + c²x² + b²x² + a²y² + c²y² + b²z² + a²z² + c²x²
= a²(x²+y²+z²) + b²(x²+y²+z²) + c²(x²+y²+z²)
= (a²+b²+c²)(x²+y²+z²) (đpcm)

b) Đặt x = b; y = c; z = a, ta có:
\(\left(ay+bz+cx\right)^2+\left(az-by\right)^2+\left(bx-cz\right)^2+\left(cy-ax\right)^2\)
= a²y² + b²z² + c²x² + 2aybz + 2bzcx + 2aycx + a²z² - 2azby + b²y² + b²x² - 2bxcz + c²z² + c²y² - 2cyax + a²x²
= a²y² + b²z² + c²x² + a²z² + b²y² + b²x² + c²z² + c²y² + a²x²
= (a²+b²+c²)(x²+y²+z²)
Thay b = x, c = y, a = z, ta có:
(a²+b²+c²)(x²+y²+z²) = (a²+b²+c²)² (đpcm)

Phương Ann Nhã Doanh Đinh Đức Hùng Mashiro Shiina

Nguyễn Thanh Hằng Nguyễn Huy Tú Lightning Farron

Akai Haruma Võ Đông Anh Tuấn

mấy anh chị cm cho e thêm cái : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)