Bài 6:
Ta có:
\(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\\ \Rightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\right]^2\ge\left(\dfrac{5}{4}\right)^2\forall x\in R\\ \Rightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\right]^2\ge\dfrac{25}{16}\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\\ Vậy:min_{BThuc}=\dfrac{25}{16}\Leftrightarrow x=-\dfrac{1}{2}\)
Câu 5:
P(x)=x3-x+5
Đặt P(x)=0
=>\(x^3-x+5=0\)
=>\(x^3-x=-5\)
=>\(x\left(x^2-1\right)=-5\)
=>\(x\left(x^2-1\right)=1\cdot\left(-5\right)=-1\cdot5=\left(-5\right)\cdot1=5\cdot\left(-1\right)\)
=>\(\left(x;x^2-1\right)\in\left\{\left(1;-5\right);\left(-1;5\right);\left(-5;1\right);\left(5;-1\right)\right\}\)
=>\(\left(x,x^2\right)\in\left\{\left(1;-4\right);\left(-1;6\right);\left(-5;2\right);\left(5;0\right)\right\}\)
=>\(x\in\varnothing\)
=>Đa thức \(P\left(x\right)=x^3-x+5\) không có nghiệm nguyên
