Question

[Lưu Vân Dung]

1. Cho (a^2 + b^2) * (x^2 + y^2) = (ax + by) với x,y khác 0. Cmr: a/x = b/y

2. Cho a^2 + b^2 + c^2 = ab+bc+ca. Cmr: a=b=c

- Giúp mk nha!!!

Answer 1

[Phương An]

a) https://hoc24.vn/hoi-dap/question/398481.html

b)

a² + b² + c² = ab + ac + bc

<=> 2a² + 2b² + 2c² = 2ac + 2ab + 2bc

<=> (a² - 2ac + c²) + (a² - 2ab + b²) + (b² - 2bc + c²) = 0

<=> (a - b)² + (a - c)² + (b - c)² = 0

<=> a = b = c

Answer 2

[Phạm Phương Anh]

1. Ta có:

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

=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

=> \(a^2y^2+b^2x^2=2axby\)

=> \(a^2y^2+b^2x^2-2axby=0\)

=> \(a^2y^2+b^2x^2-2aybx=0\)

=> \(\left(ay-bx\right)^2=0\)

Mà \(\left(ay-bx\right)^2\ge0\)

Dấu '' = '' xảy ra \(\Leftrightarrow\) \(ay-bx=0\)

\(\Leftrightarrow\) \(ay=bx\)

\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)

2. Ta có:

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

=> \(2a^2+2b^2+2c^2=2ab+2bc+2ac\)

=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

=> \(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Ta thấy:

\(\left(a-b\right)^2\ge0\); \(\left(a-c\right)^2\ge0\); \(\left(b-c\right)^2\ge0\)

=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)

Mà \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Dấu '' = '' xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\)

\(\Leftrightarrow\) a = b = c