a: Bảng giá trị:
| x | -2 | -1 | 0 | 1 | 2 |
| \(y=\dfrac{1}{2}x^2\) | 2 | 1/2 | 0 | 1/2 | 2 |
Vẽ đồ thị:
b: Phương trình hoành độ giao điểm là:
\(\dfrac{1}{4}x^2=\dfrac{1}{2}x+m^2\)
=>\(x^2=2x+4m^2\)
=>\(x^2-2x-4m^2=0\)
\(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(-4m^2\right)=16m^2+4>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
=>(P) luôn cắt (d) tại hai điểm phân biệt có hoành độ là:
\(\left[{}\begin{matrix}x=\dfrac{2-\sqrt{16m^2+4}}{2}=\dfrac{2-2\sqrt{4m^2+1}}{2}=1-\sqrt{4m^2+1}\\x=\dfrac{2+\sqrt{16m^2+4}}{2}=1+\sqrt{4m^2+1}\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=-4m^2\end{matrix}\right.\)
\(y_1-y_2+x_1^2-3x_2^2=-2\)
=>\(\dfrac{1}{4}x_1^2-\dfrac{1}{4}x_2^2+x_1^2-3x_2^2=-2\)
=>\(\dfrac{5}{4}x_1^2-\dfrac{13}{4}x_2^2=-2\)
=>\(5x_1^2-13x_2^2=-8\)
=>\(\left[{}\begin{matrix}5\cdot\left(1+\sqrt{4m^2+1}\right)^2-13\left(1-\sqrt{4m^2+1}\right)^2=-8\left(1\right)\\5\left(1-\sqrt{4m^2+1}\right)^2-13\cdot\left(1+\sqrt{4m^2+1}\right)^2=-8\left(2\right)\end{matrix}\right.\)
(1): \(5\cdot\left(1+\sqrt{4m^2+1}\right)^2-13\left(1-\sqrt{4m^2+1}\right)^2=-8\)
=>\(5\cdot\left(1+4m^2+1+2\sqrt{4m^2+1}\right)-13\left(1+4m^2+1-2\sqrt{4m^2+1}\right)=-8\)
=>\(5\left(2+4m^2+2\sqrt{4m^2+1}\right)-13\left(2+4m^2-2\sqrt{4m^2+1}\right)=-8\)
=>\(10+20m^2+10\sqrt{4m^2+1}-26-52m^2+26\sqrt{4m^2+1}=-8\)
=>\(36\sqrt{4m^2+1}=-8-10+26+32m^2\)
=>\(36\sqrt{4m^2+1}=32m^2+8\)
=>\(36\sqrt{4m^2+1}=8\left(4m^2+1\right)\)
=>\(9\sqrt{4m^2+1}=2\left(4m^2+1\right)\)
=>\(\sqrt{4m^2+1}\left(9-2\sqrt{4m^2+1}\right)=0\)
=>\(9-2\sqrt{4m^2+1}=0\)
=>\(\sqrt{4m^2+1}=4,5\)
=>\(4m^2+1=4,5^2=20,25\)
=>\(4m^2=19,25\)
=>\(m^2=\dfrac{77}{16}\)
=>\(m=\pm\dfrac{\sqrt{77}}{4}\)
(2):
\(5\left(1-\sqrt{4m^2+1}\right)^2-13\cdot\left(1+\sqrt{4m^2+1}\right)^2=-8\)
=>\(5\left(2+4m^2-2\sqrt{4m^2+1}\right)-13\left(2+4m^2+2\sqrt{4m^2+1}\right)=-8\)
=>\(10+20m^2-10\sqrt{4m^2+1}-26-52m^2-26\sqrt{4m^2+1}=-8\)
=>\(-36\sqrt{4m^2+1}-32m^2-16=-8\)
=>\(36\sqrt{4m^2+1}+32m^2+16=8\)
=>\(36\sqrt{4m^2+1}=-32m^2-8\)
=>Loại