cho pt x ²-x+m+1=0 Tìm m sao cho x1 ²+x1.x2+3×2=7 22/09/2021 Bởi Eden cho pt x ²-x+m+1=0 Tìm m sao cho x1 ²+x1.x2+3×2=7
Đáp án: m=-7 Giải thích các bước giải: Để phương trình có 2 nghiệm ⇒Δ≥0 \(\begin{array}{l} \to 1 – 4m – 4 \ge 0\\ \to – 3 \ge 4m\\ \to m \le – \dfrac{3}{4}\\ \to \left[ \begin{array}{l}x = \dfrac{{1 + \sqrt { – 3 – 4m} }}{2}\\x = \dfrac{{1 – \sqrt { – 3 – 4m} }}{2}\end{array} \right.\\Có:{x_1}^2 + {x_1}{x_2} + 3{x_2} = 7\\ \to {\left( {\dfrac{{1 + \sqrt { – 3 – 4m} }}{2}} \right)^2} + m + 1 + 3\left( {\dfrac{{1 – \sqrt { – 3 – 4m} }}{2}} \right) = 7\\ \to \dfrac{{1 + 2\sqrt { – 3 – 4m} – 3 – 4m + 4m + 4 + 6\left( {1 – \sqrt { – 3 – 4m} } \right) – 28}}{4} = 0\\ \to 2\sqrt { – 3 – 4m} + 2 + 6 – 6\sqrt { – 3 – 4m} – 28 = 0\\ \to 4\sqrt { – 3 – 4m} = 20\\ \to \sqrt { – 3 – 4m} = 5\\ \to – 3 – 4m = 25\\ \to 4m = – 28\\ \to x = – 7\left( {TM} \right)\end{array}\) Bình luận
Đáp án:
m=-7
Giải thích các bước giải:
Để phương trình có 2 nghiệm
⇒Δ≥0
\(\begin{array}{l}
\to 1 – 4m – 4 \ge 0\\
\to – 3 \ge 4m\\
\to m \le – \dfrac{3}{4}\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt { – 3 – 4m} }}{2}\\
x = \dfrac{{1 – \sqrt { – 3 – 4m} }}{2}
\end{array} \right.\\
Có:{x_1}^2 + {x_1}{x_2} + 3{x_2} = 7\\
\to {\left( {\dfrac{{1 + \sqrt { – 3 – 4m} }}{2}} \right)^2} + m + 1 + 3\left( {\dfrac{{1 – \sqrt { – 3 – 4m} }}{2}} \right) = 7\\
\to \dfrac{{1 + 2\sqrt { – 3 – 4m} – 3 – 4m + 4m + 4 + 6\left( {1 – \sqrt { – 3 – 4m} } \right) – 28}}{4} = 0\\
\to 2\sqrt { – 3 – 4m} + 2 + 6 – 6\sqrt { – 3 – 4m} – 28 = 0\\
\to 4\sqrt { – 3 – 4m} = 20\\
\to \sqrt { – 3 – 4m} = 5\\
\to – 3 – 4m = 25\\
\to 4m = – 28\\
\to x = – 7\left( {TM} \right)
\end{array}\)