Question

[Nguyễn Trần Tuấn Anh]

Answer 1

[Yen Nhi]

a.

Đa thức bậc hai cần tìm có dạng là:

\(f\left(x\right)=ax^2+bx+c\left(ĐK:a\ne0\right)\)

Có: \(f\left(x-1\right)=a\left(x-1\right)^2+b\left(x-1\right)+c\)

\(f\left(x\right)-f\left(x-1\right)=2ax-a+b=x\)

\(\Rightarrow\left\{{}\begin{matrix}2a=1\\b-a=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{2}\end{matrix}\right.\)

Vậy đa thức cần tìm là: \(f\left(x\right)=\dfrac{1}{2}x^2+\dfrac{1}{2}x+c\) (\(c\) là hằng số tuỳ í.)

Áp dụng vào, ta có:

Trường hợp: \(x=1\Rightarrow1=f\left(1\right)-f\left(0\right)\)

Trường hợp: \(x=2\Rightarrow1=f\left(2\right)-f\left(1\right)\)

...

Trường hợp: \(x=n\Rightarrow n=f\left(n\right)-f\left(n-1\right)\)

\(\Rightarrow S=1+2+3+...+n=f\left(n\right)-f\left(0\right)=\dfrac{n^2}{2}+\dfrac{n}{2}+c-c=\dfrac{n\left(n+1\right)}{2}\)

b. \(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}\)

\(\Leftrightarrow\dfrac{2abz-3acy}{a^2}=\dfrac{6bcx-2abz}{4b^2}=\dfrac{3acy-6bcx}{9c^2}=\dfrac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)

\(\Rightarrow2bz-3cy=0\)

\(\Rightarrow\dfrac{z}{3c}=\dfrac{y}{2b}\) (*)

\(\Rightarrow3cx-az=0\)

\(\Rightarrow\dfrac{x}{a}=\dfrac{z}{3c}\) (**)

Từ (*)(**)\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c}\)