Ta có : \(P\left(\dfrac{1}{2}\right)=a\left(\dfrac{1}{2}\right)^2+2.\dfrac{1}{2}+1=1\)

=> \(\dfrac{1}{4}a+1+1=1\)

\(\dfrac{1}{4}a=-1\)

\(a=-1:\dfrac{1}{4}\)

a=-4

Ta có: \(P\left(x\right)=ax^2+2x+1\)

⇒ \(P\left(\dfrac{1}{2}\right)=a.\left(\dfrac{1}{2}\right)^2+2.\dfrac{1}{2}+1=1\)

\(=\)  \(a.\dfrac{1}{4}+1+1\) \(=\dfrac{a}{4}+1:1\)

\(=\dfrac{a}{4}+1\)

⇒ \(\dfrac{a}{4}+1\)⇔ \(\dfrac{a}{4}-1\) ⇔ \(a=-4\)

Vậy hệ số a là \(-4\)

Ta có : A(x) = \(ax^2+5x-3\)

\(\Rightarrow A\left(\dfrac{1}{2}\right)=a.\left(\dfrac{1}{2}\right)^2+5.\dfrac{1}{2}-3\)

\(=a.\dfrac{1}{2}+\dfrac{5}{2}-3=\dfrac{a}{4}+\dfrac{5-2.3}{2}\)

\(=\dfrac{a}{4}-\dfrac{1}{2}\)

A(x) có nghiệm là \(\dfrac{1}{2}\)

\(\Leftrightarrow A\left(\dfrac{1}{2}\right)=0\Leftrightarrow\dfrac{a}{4}-\dfrac{1}{2}=0\)

\(\Rightarrow\dfrac{a}{4}=\dfrac{1}{2}\Rightarrow a=\dfrac{4}{2}=2\)

Vậy a = 2 .

a.\(\dfrac{1}{2}\)²+2.\(\dfrac{1}{2}\)+1=1

a.\(\dfrac{1}{4}\)+2.\(\dfrac{1}{2}\)+1=1

a.\(\dfrac{1}{4}\)+1+1=1

a.\(\dfrac{1}{4}\)+2=1

a.\(\dfrac{1}{4}\)=1-2

a.\(\dfrac{1}{4}\)=-1

a= -1:\(\dfrac{1}{4}\)

a= -4

Vì h(x) có nghiệm là 1

=> h(1)=0

=> a+5-4=0

<=> a+1=0<=> a=-1

Đa thức có nghiệm là `1 =>x=1` thỏa mãn: `a.1^2+5.1-4=0`

`<=>a+1=0`

`<=>a=-1`

Vì h(x) có nghiệm là 1

=> h(1)=0

=> a+5-4=0

<=> a+1=0<=> a=-1

a: Bậc là 2

Hệ số cao nhất là 2

Hệ số tự do là -12

b: M+N

=2x^2+5x-12+x^2-8x-1

=3x^2-3x-13

Cho P(\(\frac{1}{2}\)) = 1

⇒ P(\(\frac{1}{2}\)) = a.(\(\frac{1}{2}\))^2 + 2.(\(\frac{1}{2}\)) + 1 = 1

⇒ a.\(\frac{1}{4}\) + 1 + 1 = 1

⇒ \(\frac{a}{4}\) + 1 = 0

⇒ \(\frac{a}{4}=0\)

⇒ a = 0

Vậy hệ số a = 0