Bài 1: xét dấu
a) f(x)= 8×2 – x +7
B) f(x) = 5×2+3x+2
C) f(x)= x2+12x-7
D)f(x)= x2+12x-7
Bài 1: xét dấu a) f(x)= 8×2 – x +7 B) f(x) = 5×2+3x+2 C) f(x)= x2+12x-7 D)f(x)= x2+12x-7
By Melanie
By Melanie
Bài 1: xét dấu
a) f(x)= 8×2 – x +7
B) f(x) = 5×2+3x+2
C) f(x)= x2+12x-7
D)f(x)= x2+12x-7
Đáp án:
c) \(f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – 6 – \sqrt {43} } \right) \cup \left( { – 6 + \sqrt {43} ; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)f\left( x \right) = 8{x^2} – x + 7 = 8{x^2} – 2.2\sqrt 2 x.\dfrac{7}{{2.2\sqrt 2 }} + \dfrac{{49}}{{32}} + \dfrac{{175}}{{32}}\\
= {\left( {2\sqrt 2 x – \dfrac{7}{{2.2\sqrt 2 }}} \right)^2} + \dfrac{{175}}{{32}} > 0\forall x\\
b)f\left( x \right) = 5{x^2} + 3x + 2 = {\left( {x\sqrt 5 } \right)^2} + 2.x\sqrt 5 .\dfrac{1}{{\sqrt 5 }} + \dfrac{1}{5} + \dfrac{9}{5}\\
= {\left( {x\sqrt 5 + \dfrac{1}{{\sqrt 5 }}} \right)^2} + \dfrac{9}{5} > 0\forall x\\
c)f\left( x \right) = {x^2} + 12x – 7\\
\to {x^2} + 12x – 7 = 0\\
\to \left[ \begin{array}{l}
x = – 6 + \sqrt {43} \\
x = – 6 – \sqrt {43}
\end{array} \right.
\end{array}\)
x -∞ \( – 6 – \sqrt {43} \) \( – 6 + \sqrt {43} \) +∞
f(x) + 0 – 0 +
\(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – 6 – \sqrt {43} } \right) \cup \left( { – 6 + \sqrt {43} ; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { – 6 – \sqrt {43} ; – 6 + \sqrt {43} } \right)
\end{array}\)
a) $f(x)=8x^2-x+7$
Có: $a=8>0; Δ=-223<0$
$⇒x∈R$
b) $f(x)=5x^2+3x+2$
Có: $a=5>0; Δ=-31<0$
$⇒x∈R$
c) $f(x)=x^2+12x-7$
Có: $a=1>0; Δ=172>0$
Cho $x^2+12x-7=0$
$⇒x1=-6+√43; x2=-6-√43$
BXD:
$x$ $-∞$ $-6-√43$ $-6+√43$ $+∞$
$f(x)$ $+$ $0$ $-$ $0$ $+$