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a) \(a:b:c=\left(-1\right):3:\left(-4\right)\Rightarrow-a=\dfrac{b}{3}=-\dfrac{c}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}b=-3a\\c=4a\end{matrix}\right.\)
\(\dfrac{1}{2}f\left(2\right)=-2\)
\(\Rightarrow\dfrac{1}{2}.\left(4a+2b+c\right)=-2\)
\(\Rightarrow2a+b+\dfrac{c}{2}=-2\)
\(\Rightarrow2a-3a+\dfrac{4a}{2}=-2\)
\(\Rightarrow a=-2\)
\(\Rightarrow\left\{{}\begin{matrix}b=-3a=-3.\left(-2\right)=6\\c=4a=4.\left(-2\right)=-8\end{matrix}\right.\).
b) \(f\left(x\right)=h\left(x\right)+11x^2+6x+2\)
\(\Rightarrow-2x^2+6x-8=h\left(x\right)+11x^2+6x+2\)
\(\Rightarrow h\left(x\right)=-13x^2-10\)
\(\Rightarrow h\left(x\right)=-\left(13x^2+10\right)\le-\left(13+10\right)=-23\)
\(h\left(x\right)=-23\Leftrightarrow x=0\)
-Vậy \(h\left(x\right)_{max}=-23\)
Lời giải:
$f(1)=a+b+c=6$
$f(2)=4a+2b+c=16$
$f(12)-f(-9)=(144a+12b+c)-(81a-9b+c)$
$=63a+21b=21(3a+b)$
$=21[(4a+2b+c)-(a+b+c)]=21(16-6)=21.10=210$
a) Thay x = 1 vào đa thức F(x), ta có:
F(1) = a.12 + b.1 + c = a+ b + c
Mà a + b + c = 0
Do đó, F(1) = 0. Như vậy x = 1 là một nghiệm của F(x)
b) Ta có: Đa thức 2x2 – 5x + 3 có a = 2 ; b = -5; c = 3 nên a + b + c = 2 + (-5) + 3 = 0
Do đó, đa thức có 1 nghiệm là x = 1
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{121}\right)\)
\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{120}{121}\)
\(-A=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot10\cdot12}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot11\cdot11}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot...\cdot11\right)\left(2\cdot3\cdot4\cdot...\cdot11\right)}\)
\(-A=\frac{1\cdot12}{11\cdot2}=\frac{6}{11}\)
\(A=-\frac{6}{11}\)
\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(B=1-\frac{1}{38}=\frac{37}{38}\)
\(P\left(x\right)=ax^2+bx+c\)
Suy ra \(P\left(-1\right)=a-b+c=-1\)\(\Rightarrow a-\left(b-c\right)=-1\)(1)
\(P\left(-\frac{1}{3}\right)=\frac{a}{9}-\frac{b}{3}+c=\frac{17}{9}\); \(P\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c=\frac{17}{4}\)
\(\Rightarrow\hept{\begin{cases}\frac{a-3b+9c}{9}=\frac{17}{9}\\\frac{a+2b+4c}{4}=\frac{17}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}a-3b+9c=17\\a+2b+4c=17\end{cases}}\)(2)
\(\Rightarrow a-3b+9c=a+2b+4c\Leftrightarrow a-3b+9c-a-2b-4c=0\)
\(\Leftrightarrow-5b+5c=0\Leftrightarrow-5\left(b-c\right)=0\Rightarrow b-c=0\)
Thay b - c = 0 vào (1) ta có: \(a-0=-1\Leftrightarrow a=-1\)
Thay a=-1 vào (2) \(\Rightarrow\hept{\begin{cases}-1-\left(3b-9c\right)=17\\-1+2b+4c=17\end{cases}}\Leftrightarrow\hept{\begin{cases}3b-9c=-18\\2b+4c=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}6b-18c=-36\\6b+12c=54\end{cases}}\Rightarrow6b+12c-6b+18c=54+36=90\)
\(\Leftrightarrow30c=90\Leftrightarrow c=3.\)Do \(b-c=0\Leftrightarrow b=c\Rightarrow b=3\)
\(\Rightarrow a+b+c=-1+2.3=5.\)Vậy...........