mày bị điên đứa nào thích thì mà đứa nào chơi truy kích cho tao nick

Đặt \(f\left(x\right)=\left(a_1x-b_1\right)^2+...+\left(a_nx-b_n\right)^2\)

\(\Rightarrow f\left(x\right)\ge0\) với mọi x

Mặt khác : \(f\left(x\right)=\left(a_1^2+...+a_n^2\right)x^2-2\left(a_1b_1+...+a_nb_n\right)x+\left(b_1^2+...+b_n^2\right)\)

\(\Rightarrow\Delta'\le0\)

\(\Rightarrow\left(a_1b_1+...+a_nb_n\right)^2\le\left(a_1^2+...+a_n^2\right)\left(b_1^2+...+b_n^2\right)\)

\(\Rightarrow\left|a_1b_1+...+a_nb_n\right|\le\sqrt{\left(a_1^2+...+a_n^2\right)\left(b_{1^{ }}^2+...+b_n^2\right)}\)

Áp dụng bđt bunhia copski, ta có \(\left(a_1b_1+...+a_nb_n\right)^2\le\left(a_1^2+...+a_n^2\right)\left(b_1^2+...+b_2^2\right)\Leftrightarrow\sqrt{\left(a_1b_1+...+a_nb_n\right)^2}\le\sqrt{\left(a_1^2+...+a_n^2\right)\left(b_1^2+...+b_2^2\right)}\Leftrightarrow\left|a_1b_1+...+a_nb_n\right|\le\sqrt{\left(a_1^2+...+a_n^2\right)\left(b_1^2+...+b_2^2\right)}\)

Dấu bằng xảy ra khi \(\dfrac{a_1}{b_1}=...=\dfrac{a_n}{b_n}\)

Vậy \(\left|a_1b_1+...+a_nb_n\right|\le\sqrt{\left(a_1^2+...+a_n^2\right)\left(b_1^2+...+b_2^2\right)}\)

a) Đặt \(d=\left(a_1,a_2,...,a_n\right)\Rightarrow\left\{{}\begin{matrix}a_1=dx_1\\a_2=dx_2\\...\\a_n=dx_n\end{matrix}\right.\) (với \(\left(x_1,x_2,...,x_n\right)=1\)).

Ta có \(A_i=\dfrac{A}{a_i}=\dfrac{d^nx_1x_2...x_n}{dx_i}=d^{n-1}\dfrac{x_1x_2...x_n}{x_i}=d^{n-1}B_i\forall i\in\overline{1,n}\).

Từ đó \(\left[A_1,A_2,...,A_n\right]=d^{n-1}\left[B_1,B_2,...,B_n\right]\).

Mặt khác do \(\left(x_1,x_2,...,x_n\right)=1\Rightarrow\left[B_1,B_2,...B_n\right]=x_1x_2...x_n\).

Vậy \(\left(a_1,a_2,...,a_n\right)\left[A_1,A_2,...,A_n\right]=d.d^{n-1}x_1x_2...x_n=d^nx_1x_2...x_n=A\).

cho toán lớp 10 ....Bố nó hiểu