\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)

Ta có: f(0)=1

<=> ax² +bx+c=1

<=> c=1

          f(1)=0

<=>ax² +bx+c=0

<=> a+b+c=0

mà c=1

=>a+b=-1(1)

      f(-1)=10

<=> ax² +bx +c=10

<=>a-b+c=10

mà c=1

=>a-b=9(2)

Lấy (1) trừ (2) ta được (a+b)-(a-b)=-1-9

                           <=> 2b=-10

                           <=> b=-5

                           =>a=4

Vậy a=4,b=-5,c=1

Nhớ k đúng cho mik

Lời giải:
a. 

$f(-1)=a-b+c$

$f(-4)=16a-4b+c$

$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$

$\Rightarrow f(-4)=6f(-1)$

$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)

b.

$f(-2)=4a-2b+c$

$f(3)=9a+3b+c$

$\Rightarrow f(-2)+f(3)=13a+b+2c=0$

$\Rightarrow f(-2)=-f(3)$

$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)

a. 

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2
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\(f\left(-1\right)=-a+b-c+d=2\)

\(f\left(0\right)=d=1\)

\(f\left(\frac{1}{2}\right)=\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c+d=3\)

\(f\left(1\right)=a+b+c+d=7\)

Suy ra \(\hept{\begin{cases}-a+b-c=1\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}2b=7\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{7}{2}\\c=\frac{13}{6}\end{cases}}\)

\(\left\{{}\begin{matrix}f\left(0\right)=5\Rightarrow0+0+5\Rightarrow c=5\\f\left(1\right)=0\Rightarrow a+b+5=0\\f\left(5\right)=0\Rightarrow25a+5b+5=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\right.\)

tu (3) => b =-1-5a

tu (2) => a-1-5a+5 =0 => a =1 ;b =-6

y =x^2 -6x +5

y(-1) =1 +6 +5 khac 3 => loai

y(-1/2) =1/4 -6/2 +5 =1/4 +2 = 9/4 nhan

Q(1/2;9/4) thuoc dths

Gửi bạn

Ta có \(\left\{{}\begin{matrix}f\left(0\right)=5\\f\left(-1\right)=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=5\\-a+b=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=5\\a=3\end{matrix}\right.\)

Vậy hàm số \(y=ax+b=3x+5\) 

Ta có: f(0)=5

nên b=5

hay y=ax+5

Thay x=-1 và y=2 vào y=ax+5, ta được:

\(-a+5=2\)

hay a=3

\(f\left(x_1\right)=ax_1\) ; \(f\left(x_2\right)=ax_2\) ; \(f\left(x_1x_2\right)=ax_1x_2\)

Để \(f\left(x_1\right)f\left(x_2\right)=f\left(x_1x_2\right)\)

\(\Leftrightarrow ax_1.ax_2=ax_1x_2\)

\(\Leftrightarrow a^2x_1x_2=ax_1x_2\)

\(\Leftrightarrow a^2=a\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=1\end{matrix}\right.\)

Vậy \(a=1\)