Question

cm 2x^2 +7x+11 vô no

Answer 1

Ta có:

`2x^2+7x+11`

`=2(x^2+7/2x+11/2)`

`=2[(x^2+2*x*7/4+49/16)+39/16]`

`=2(x+7/4)^2+39/8`

Vì: `(x+7/4)^2>=0` với mọi x

`=>2(x+7/4)^2>=0` với mọi x

`=>2(x+7/4)^2+39/8>=39/8>0` với mọi x

`=>` Vô nghiệm 

Answer 2

\(2x^2+7x+11=0\)

\(\Leftrightarrow2\left(x^2+\dfrac{7}{2}x+\dfrac{11}{2}\right)=0\)

\(\Leftrightarrow x^2+\dfrac{7}{2}x+\dfrac{11}{2}=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{7}{4}+\left(\dfrac{7}{4}\right)^2+\dfrac{39}{16}=0\)

\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2+\dfrac{39}{16}=0\)

\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2=-\dfrac{39}{16}\)

\(\Rightarrow\)Vô lý vì \(\left(x+\dfrac{7}{4}\right)^2\ge0\forall x\)

Vậy...