* x^2 -2(m+1)x-3=0. tìm các giá trị của m sao cho /x1/-/x2/=5 và x1
* x^2 -2(m+1)x-3=0. tìm các giá trị của m sao cho /x1/-/x2/=5 và x1
By Delilah
By Delilah
Đáp án:
$\begin{array}{l}
1)\\
{x^2} – 2\left( {m + 1} \right)x – 3 = 0\\
\Rightarrow \Delta ‘ > 0\\
\Rightarrow {\left( {m + 1} \right)^2} + 3 > 0\left( {luon\,dung} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = – 3
\end{array} \right.\\
\left| {{x_1}} \right| – \left| {{x_2}} \right| = 5\\
\Rightarrow x_1^2 – 2\left| {{x_1}} \right|.\left| {{x_2}} \right| + x_2^2 = 25\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} – 2{x_1}{x_2} – 2.\left| { – 3} \right| = 25\\
\Rightarrow 4{\left( {m + 1} \right)^2} – 2.\left( { – 3} \right) – 2.3 = 25\\
\Rightarrow {\left( {m + 1} \right)^2} = \dfrac{{25}}{4}\\
\Rightarrow \left[ \begin{array}{l}
m = – 1 + \dfrac{5}{2} = \dfrac{3}{2}\\
m = – 1 – \dfrac{5}{2} = \dfrac{{ – 7}}{2}
\end{array} \right.\\
2)\\
{x^2} – 6x + m = 0\\
\Rightarrow \Delta ‘ > 0\\
\Rightarrow 9 – m > 0\\
\Rightarrow m < 9\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1}{x_2} = m
\end{array} \right.\\
x_1^2 – x_2^2 = 12\\
\Rightarrow \left( {{x_1} + {x_2}} \right)\left( {{x_1} – {x_2}} \right) = 12\\
\Rightarrow {x_1} – {x_2} = 2\\
\Rightarrow {\left( {{x_1} – {x_2}} \right)^2} = 4\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} – 4{x_1}{x_2} = 4\\
\Rightarrow 36 – 4.m = 4\\
\Rightarrow m = 8\left( {tmdk} \right)\\
3)\\
\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 4} \right)\left( {x + 5} \right) = 10\\
\Rightarrow \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right)\left( {x + 4} \right) = 10\\
\Rightarrow \left( {{x^2} + 6x + 5} \right)\left( {{x^2} + 6x + 8} \right) = 10\\
Dat:{x^2} + 6x + 5 = a\\
\Rightarrow a.\left( {a + 3} \right) = 10\\
\Rightarrow {a^2} + 3a – 10 = 0\\
\Rightarrow \left( {a – 2} \right)\left( {a + 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a – 2 = 0\\
a + 5 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 6x + 3 = 0\\
{x^2} + 6x + 10 = 0\left( {vô\,nghiệm} \right)
\end{array} \right.\\
\Rightarrow {x^2} + 6x + 9 – 6 = 0\\
\Rightarrow {\left( {x + 3} \right)^2} = 6\\
\Rightarrow \left[ \begin{array}{l}
x = – 3 + \sqrt 6 \\
x = – 3 – \sqrt 6
\end{array} \right.
\end{array}$