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f(0) = 1
\(\Rightarrow\) a.02 + b.0 + c = 1
\(\Rightarrow\) c = 1
Vậy hệ số a = 0; b = 0; c = 1
f(1) = 2
\(\Rightarrow\) a.12 + b.1 + c = 2
\(\Rightarrow\) a + b + c = 2
Vậy hệ số a = 1; b = 1; c = 1
f(2) = 4
\(\Rightarrow\) a.22 + b.2 + c = 4
\(\Rightarrow\) 4a + 2b + c = 4
Vậy hệ số a = 4; b = 2; c = 1
Chúc bn học tốt! (chắc vậy :D)
\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)
\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)
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
\(f\left(-1\right)=-a+b-c+d=2\)
\(f\left(0\right)=d=1\)
\(f\left(\frac{1}{2}\right)=\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c+d=3\)
\(f\left(1\right)=a+b+c+d=7\)
Suy ra \(\hept{\begin{cases}-a+b-c=1\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}2b=7\\\frac{1}{8}a+\frac{1}{4}b+\frac{1}{2}c=2\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{7}{2}\\c=\frac{13}{6}\end{cases}}\)
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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⇒f(−4)−6f(−1)=16a−4b+c−6(a−b+c)=10a+2b−5c=0
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⇒f(−1)f(−4)=f(−1).6f(−1)=6[f(−1)]
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⇒f(−2)f(3)=−[f(3)]
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\(\left\{{}\begin{matrix}f\left(0\right)=5\Rightarrow0+0+5\Rightarrow c=5\\f\left(1\right)=0\Rightarrow a+b+5=0\\f\left(5\right)=0\Rightarrow25a+5b+5=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\right.\)
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
\(f\left(-1\right)=a\left(-1\right)^2+b.\left(-1\right)+c\)
\(=a-b+c\)
\(f\left(2\right)=a.2^2+b.2+c\)
\(=4a+2b+c\)
\(\Rightarrow f\left(2\right)-2.f\left(-1\right)=\left(4a+2b+c\right)-2\left(a-b+c\right)\)
\(=2a+4b-c=0\)
\(\Rightarrow f\left(2\right)=2.f\left(-1\right)\)
\(\Rightarrow f\left(2\right)\)và \(2.f\left(-1\right)\)cùng dấu
\(\Rightarrow f\left(2\right)\)và \(f\left(-1\right)\)cùng dấu
\(\Rightarrow f\left(2\right).f\left(-1\right)\ge0\)(đpcm)
Ta có :\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c\)
\(\implies\) \(f\left(2\right)-2f\left(-1\right)=\left(4a+2b+c\right)-2.\left(a-b+c\right)\)
\(\implies\) \(f\left(2\right)=2.f\left(-1\right)\)
\(\implies\) \(f\left(-1\right).f\left(2\right)=f\left(-1\right).2f\left(-1\right)=f\left(-1\right)^2.2\) \(\geq\) \(0\)
\(\implies\) \(f\left(-1\right).f\left(2\right)\) \(\geq\) \(0\) \(\left(đpcm\right)\)
Ta có \(f\left(0\right)=1\)
\(\Rightarrow a\cdot0^2+b\cdot0+c=1\\ \Rightarrow0+0+c=1\\ \Rightarrow c=1\)
\(f\left(1\right)=0\\ \Rightarrow a\cdot1^2+b\cdot1+c=0\\ \Rightarrow a+b+c=0\\ \Rightarrow a+b=-1\left(1\right)\)
\(f\left(-1\right)=6\\ \Rightarrow a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=6\\ \Rightarrow a-b+c=6\\ \Rightarrow a-b=5\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow2a=4\\ \Rightarrow a=2\\ \Rightarrow b=-1-a=-1-2=-3\)
Vậy a = 2 ; b = -3 ; c = 1
\(f\left(x\right)=ax^2+bx+c\)
+ \(f\left(0\right)=1.\)
\(\Rightarrow f\left(0\right)=a.0^2+b.0+c=1\)
\(\Rightarrow f\left(0\right)=a.0+b.0+c=1\)
\(\Rightarrow f\left(0\right)=0+0+c=1\)
\(\Rightarrow f\left(0\right)=c=1\)
\(\Rightarrow c=1.\)
+ \(f\left(1\right)=0.\)
\(\Rightarrow f\left(1\right)=a.1^2+b.1+c=0\)
\(\Rightarrow f\left(1\right)=a.1+b.1+c=0\)
\(\Rightarrow f\left(1\right)=a+b+c=0\)
\(\Rightarrow a+b+c=0\)
Mà \(c=1\left(cmt\right).\)
\(\Rightarrow a+b+1=0\)
\(\Rightarrow a+b=0-1\)
\(\Rightarrow a+b=-1\) (1).
+ \(f\left(-1\right)=6.\)
\(\Rightarrow f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=6\)
\(\Rightarrow f\left(-1\right)=a.1+b.\left(-1\right)+c=6\)
\(\Rightarrow f\left(-1\right)=a+\left(-b\right)+c=6\)
\(\Rightarrow f\left(-1\right)=a-b+c=6\)
\(\Rightarrow a-b+c=6\)
Mà \(c=1\left(cmt\right).\)
\(\Rightarrow a-b+1=6\)
\(\Rightarrow a-b=6-1\)
\(\Rightarrow a-b=5\) (2).
Cộng theo vế (1) và (2) ta được:
\(a+b+a-b=\left(-1\right)+5\)
\(\Rightarrow2a=4\)
\(\Rightarrow a=4:2\)
\(\Rightarrow a=2.\)
+ Ta có: \(a+b=-1.\)
\(\Rightarrow2+b=-1\)
\(\Rightarrow b=\left(-1\right)-2\)
\(\Rightarrow b=-3.\)
Vậy \(a=2;b=-3;c=1.\)
Chúc bạn học tốt!