Câu hỏi của Đỗ Châu Oanh

Question

Cho (P); y=x2 và (d); y=(m+1)x-m

Tìm m để (d) cắt (P) tại 2 điểm phân biệt có hoành độ x1 và x2 thỏa mãn: y1+y2 min.

Lê Song Phương · Accepted Answer

Equation of the intersection of (P) and (d) is:

$x^2=\left(m+1\right)x-m$ $\Leftrightarrow x^2-\left(m+1\right)x+m=0$  (1)$a=1;b=-\left(m+1\right);c=m$

We can see that $a+b+c=1-\left(m+1\right)+m=0$ so the equation (1) has 2 roots: $x_1=1;x_2=\dfrac{c}{a}=m$

We have $y_1=x_1^2=1^2=1$; $y_2=x_2^2=m^2$

Thus, $y_1+y_2=1+m^2$

Because $m^2\ge0\Leftrightarrow m^2+1\ge1$ or $y_1+y_2\ge1$. "=" happens when $m=0$

In conclusion, in order to minimize the value of $y_1+y_2$, m must be equal to 0.Câu trả lời: Equation of the intersection of (P) and (d) is:

$x^2=\left(m+1\right)x-m$ $\Leftrightarrow x^2-\left(m+1\right)x+m=0$  (1)$a=1;b=-\left(m+1\right);c=m$

We can see that $a+b+c=1-\left(m+1\right)+m=0$ so the equation (1) has 2 roots: $x_1=1;x_2=\dfrac{c}{a}=m$

We have $y_1=x_1^2=1^2=1$; $y_2=x_2^2=m^2$

Thus, $y_1+y_2=1+m^2$

Because $m^2\ge0\Leftrightarrow m^2+1\ge1$ or $y_1+y_2\ge1$. "=" happens when $m=0$

In conclusion, in order to minimize the value of $y_1+y_2$, m must be equal to 0.