Giả bài sau theo các bước : B1 : Tìm ĐK để pt có 2 ` n^o ` ` x_1 ` , ` x_2 ` Tìm ĐK để pt có 2 ` n^o ` ` x_1 ` , ` x_2 ` $\left \{ {{ a \ne 0 } \atop { Δ \geq 0 }} \right.$ B2 : Áp dụng định lí vi - ét tính được ` S ` ` = `` x_1 + x_2 ` , ` P ` ` = ` ` x_1 . x

honhattam2311

2 năm trước

Giả bài sau theo các bước : B1 : Tìm ĐK để pt có 2 ` n^o ` ` x_1 ` , ` x_2 ` Tìm ĐK để pt có 2 ` n^o ` ` x_1 ` , ` x_2 ` $\left \{ {{ a \ne 0 } \atop { Δ \geq 0 }} \right.$ B2 : Áp dụng định lí vi - ét tính được ` S ` ` = `` x_1 + x_2 ` , ` P ` ` = ` ` x_1 . x_2 ` B3 : Từ ĐK ( T ) và S tính ` x_1 ; x_2 ` theo ` m ` thế vào ` P ` để tìm thử lại ĐK

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thaovytran2008

Đáp án: $m\in\{2,-14\}$

Giải thích các bước giải:

Ta có $x^2-2\sqrt{15}x+m+4=0$ là phương trình bậc $2$ ẩn $x$

$\to $Để phương trình có $2$ nghiệm

$\Delta'\ge 0$

$\to (\sqrt{15})^2-1(m+4)\ge 0$

$\to m\le 11$

Khi đó phương trình có $2$ nghiệm $x_1,x_2$ thỏa mãn:

$\begin{cases}x_1+x_2=2\sqrt{15}\\x_1x_2=m+4\end{cases}$

Để $\dfrac{x_1}{x_2}-\dfrac{x_2}{x_1}=2\sqrt{15}$

$\to\dfrac{x_1^2-x_2^2}{x_1x_2}=2\sqrt{15}$

$\to \dfrac{(x_1-x_2)(x_1+x_2)}{x_1x_2}=2\sqrt{15}$

$\to \dfrac{(x_1-x_2)\cdot 2\sqrt{15}}{m+4}=2\sqrt{15}$

$\to x_1-x_2=m+4$

$\to (x_1+x_2)-2x_2=m+4$

$\to 2\sqrt{15}-2x_2=m+4$

$\to 2x_2=2\sqrt{15}-m-4$

$\to x_2=\dfrac{2\sqrt{15}-m-4}2$

$\to x_1=2\sqrt{15}-x_2=2\sqrt{15}-\dfrac{2\sqrt{15}-m-4}2=\dfrac{2\sqrt{15}+m+4}{2}$

Mà $x_1x_2=m+4$

$\to \dfrac{2\sqrt{15}-m-4}2\cdot \dfrac{2\sqrt{15}+m+4}{2}=m+4$

$\to \dfrac{(2\sqrt{15})^2-(m+4)^2}{4}=m+4$

$\to \dfrac{60-(m+4)^2}{4}=m+4$

$\to m\in\{2,-14\}$ thỏa mãn $m\le 11$

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