Câu 5:
a: f(x)>0
=>\(2x^2-5x+2>0\)
=>(2x-1)(x-2)>0
=>\(\left[{}\begin{matrix}x>2\\x< \dfrac{1}{2}\end{matrix}\right.\)
=>Sai
b: f(x)=9-x2=(3-x)(3+x)
f(x)>0
=>(3-x)(3+x)>0
=>(x-3)(x+3)<0
=>-3<x<3
=>Đúng
c: \(\Delta=\left(\sqrt{7}-1\right)^2-4\cdot1\cdot\left(\sqrt{3}\right)=8-2\sqrt[]{7}-4\sqrt{3}< 0\)
mà a=1>0
nên f(x)>0 với mọi x
=>Đúng
d: \(f\left(x\right)=-x^2+x-\dfrac{1}{4}=-\left(x^2-x+\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2\)
Khi x<>1/2 thì \(\left(x-\dfrac{1}{2}\right)^2>0\)
=>\(-\left(x-\dfrac{1}{2}\right)^2< 0\)
=>Đúng
Câu 6:
a: \(x^2+4x+3< 0\)
=>(x+3)(x+1)<0
=>-3<x<-1
=>Đúng
b: \(x^2-6x+8>=0\)
=>(x-2)(x-4)>=0
=>\(\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\)
=>Đúng
c: \(f\left(x\right)=x^2-x+5=x^2-x+\dfrac{1}{4}+\dfrac{19}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}>=\dfrac{19}{4}>0\forall x\)
=>Sai
d: \(f\left(x\right)=-36x^2+12x-1\)
\(=-\left(36x^2-12x+1\right)\)
\(=-\left(6x-1\right)^2< =0\forall x\)
=>Đúng
giải chi tiết giúp mình với ạ 