2/
B 2 = x 2 ( 1 − x 2 ) = − x 4 + x 2 = − ( x 4 − x 2 + 1 4 ) + 1 4 = 1 4 − ( x 2 − 1 2 ) 2 ≤ 1 4 B^2=x^2(1-x^2)=-x^4+x^2=-(x^4-x^2+\frac{1}{4})+\frac{1}{4}=\frac{1}{4}-(x^2-\frac{1}{2})^2\leq \frac{1}{4} B 2 = x 2 ( 1 − x 2 ) = − x 4 + x 2 = − ( x 4 − x 2 + 4 1 ) + 4 1 = 4 1 − ( x 2 − 2 1 ) 2 ≤ 4 1
⇔ − 1 2 ≤ B ≤ 1 2 \Leftrightarrow \frac{-1}{2}\leq B\leq \frac{1}{2} ⇔ 2 − 1 ≤ B ≤ 2 1
3/
A 2 = 4 x 2 ( 5 − x 2 ) = − 4 x 4 + 20 x 2 = − ( 4 x 4 − 2.2 x 2 . 5 + 25 ) + 25 = 25 − ( 2 x 2 − 5 ) 2 ≤ 25 A^2=4x^2(5-x^2)=-4x^4+20x^2=-(4x^4-2.2x^2.5+25)+25=25-(2x^2-5)^2\leq 25 A 2 = 4 x 2 ( 5 − x 2 ) = − 4 x 4 + 2 0 x 2 = − ( 4 x 4 − 2 . 2 x 2 . 5 + 2 5 ) + 2 5 = 2 5 − ( 2 x 2 − 5 ) 2 ≤ 2 5
⇒ − 5 ≤ A ≤ 5 \Rightarrow -5\leq A\leq 5 ⇒ − 5 ≤ A ≤ 5
4/ Điều kiện:
− 1 ≤ x ≤ 1 -1\leq x\leq 1 − 1 ≤ x ≤ 1
Ta có:
A 2 = 25 − 30 x + 9 x 2 1 − x 2 A^2=\frac{25-30x+9x^2}{1-x^2} A 2 = 1 − x 2 2 5 − 3 0 x + 9 x 2
⇔ x 2 ( 9 + A 2 ) − 30 x + 25 − A 2 = 0 \Leftrightarrow x^2(9+A^2)-30x+25-A^2=0 ⇔ x 2 ( 9 + A 2 ) − 3 0 x + 2 5 − A 2 = 0
Để phương trình theo nghiệm x có nghiệm thì
Δ ′ = 1 5 2 − ( 25 − A 2 ) ( 9 + A 2 ) ≥ 0 \Delta '=15^2-(25-A^2)(9+A^2)\geq 0 Δ ′ = 1 5 2 − ( 2 5 − A 2 ) ( 9 + A 2 ) ≥ 0
⇔ A 2 ≥ 16 o r A 2 = 0 ( l ) \Leftrightarrow A^2\geq 16 or A^{2}=0(l) ⇔ A 2 ≥ 1 6 o r A 2 = 0 ( l )
⇒ A ≥ 4 \Rightarrow A\geq 4 ⇒ A ≥ 4
Last edited by a moderator: 8 Tháng mười một 2017