ý b

ta thấy\(\sqrt{\left(2x+1\right)^2+4}\ge0\forall x\)

3/4y\(^2\)-1\(\ge0\forall x\)

suy ra \(\sqrt{\left(2x+1\right)^2+4}+3\)/4y\(^2\)-1/\(\ge0\forall x,y\)

=>min a=5

dau =xảy ra <=>x=\(\dfrac{3}{2}\),y=\(\dfrac{1}{2}\)

\(f\left(x\right)=-x^2-5x-10=-\left(x^2+2x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right)+\left(\dfrac{5}{2}\right)^2-10\)

\(f\left(x\right)=-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}\)

ta thay \(-\left(x+\dfrac{5}{2}\right)^2\le0\)

\(\Rightarrow-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}\le-\dfrac{15}{4}\)

\(\Rightarrow-\left(x+\dfrac{5}{2}\right)^2-\dfrac{15}{4}< 0\)

suy ra da thuc f(x) vo nghiem

cho đa thức f(x)=\(ax^2+bx+c\)

ta có:f(0)=c\(\in\)z(1)

f(1)=a+b+c\(\in\)zmà c\(\in\)z

=>a+b\(\in\)z(2)

f(2)=4a+2b+c\(\in z\)mà c\(\in\)z

=>4a+2b\(\in\)z(3)

từ (3)(2)ta có( 4a+2b)-(a+b)=3a-b\(\in\)z

mà 3\(\in\)z=>a-b\(\in\)z(4)

từ (2)(4)=>a+b+a-b=2a\(\in\)

mà 2\(\in\)z=>a\(\in\)z(5)

=>a\(\in\)z mà a-b\(\in\)z=>b\(\in\)z(6)

từ (1)(5)(6)=>f(x) nguyên với mọi giá trị x nguyên

much