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Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(-2\right)=4a-2b+c\)
\(f\left(3\right)=9a+3b+c\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)(vì 13a+b+2c=0)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)=-\left[f\left(-2\right)\right]^2\le0\)( đpcm)
\(f\left(0\right)=5\\ \Leftrightarrow a\cdot0^2+b\cdot0+c=c=5\\\Rightarrow c=5\\ f\left(1\right)=3\\ \Leftrightarrow a\cdot1^2+b\cdot1+c=a+b+5=3\\ \Leftrightarrow a+b=-2\\ \Leftrightarrow2a+2b=-4\\ f\left(-2\right)=4\\ \Leftrightarrow a\cdot\left(-2\right)^2+b\cdot\left(-2\right)+c=4a-2b+5=4\\ \Leftrightarrow4a-2b=-1\\ 2a+2b+4a-2b=-4+\left(-1\right)\\ \Leftrightarrow6a=-5\\ \Leftrightarrow a=\dfrac{-5}{6}\\ a+b=-2\\ \Leftrightarrow\dfrac{-5}{6}+b=-2\\ \Leftrightarrow b=\dfrac{-7}{6}\)
c) +)Điểm A ( 1;9) => x = 1 ; y = 9
Thay x = 1 vào y = 4x+5 , ta có:
y = 4.1+5
y = 4+5
y = 9
Vậy điểm A ( 1;9 ) thuộc đồ thị hàm số y = 4x +5
+) Điểm B ( -2;3 ) => x = -2 ; y = 3
Thay x = -2 vào y = 4x +5 , ta có:
y = 4.(-2) + 5
y = (-8) + 5
y = (-3)
Vậy điểm B ( -2;3) không thuộc đồ thị hàm số y = 4x+5
....Các câu khác tương tự....> . <...
a,\(2x^2-8x=0\)
\(2x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b,\(B\left(x\right)=\left(2x^2-8x\right)-\left(3x+2x^2\right)\)
\(=2x^2-8x-3x-2x^2\)
=\(-11x\)
c,\(-11x=0\)
\(\Rightarrow x=0\)
\(A\left(x\right)=2x^2-8x\)
\(\Rightarrow2x^2-8x=0\)
\(\Rightarrow x\left(2x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=8\Rightarrow x=4\end{matrix}\right.\)
\(B\left(x\right)=-3x+2x^2\)
\(B\left(x\right)=2x^2-3x\)
\(2x^2-3x=0\)
\(\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
Ta có:+)f(2017)=ax^3 + bx +5=5
x(ax^2 + bx)=0
=>ax^2 + bx=0(do x=-2017)
+)f(-2017)=ax^3 + bx +5
=x(ax^2 +bx)+5
=x.0+5=0+5=5
a) Giải:
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c\\f\left(3\right)=a.3^2+b.3+c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(-2\right)=4a-2b+c\\f\left(3\right)=9a+3b+c\end{matrix}\right.\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=\left(4a-2b+c\right)+\left(9a+3b+c\right)\)
\(=\left(4a+9a\right)+\left(-2b+3b\right)+\left(c+c\right)\)
\(=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)=-\left[f\left(3\right)\right]^2\le0\)
Vậy \(f\left(-2\right).f\left(3\right)\le0\) (Đpcm)
b) Sửa đề:
Biết \(5a+b+2c=0\)
Giải:
Ta có:
\(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=a.2^2+b.2+c=4a+2b+c\\f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\end{matrix}\right.\)
\(\Rightarrow f\left(2\right)+f\left(-1\right)=\left(a-b+c\right)+\left(4a+2b+c\right)\)
\(=\left(4a+a\right)+\left(-b+2b\right)+\left(c+c\right)\)
\(=5a+b+2c=0\)
\(\Rightarrow f\left(2\right)=-f\left(-1\right)\)
\(\Rightarrow f\left(2\right).f\left(-1\right)=-\left[f\left(-1\right)\right]^2\le0\)
Vậy \(f\left(2\right).f\left(-1\right)\le0\) (Đpcm)
Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\hept{\begin{cases}f\left(-3\right)=9a-3b+c\\f\left(4\right)=16a+4a+c\end{cases}}\) \(\Rightarrow f\left(-3\right)+f\left(4\right)=25a+b+2c=0\)
\(\Rightarrow f\left(-3\right)=-f\left(4\right)\)
Khi đó: \(f\left(-3\right)\cdot f\left(4\right)=-f\left(4\right)\cdot f\left(4\right)=-\left[f\left(4\right)\right]^2< 0\)
Đề bài bị sai rồi phần đpcm phải là "\(\le\)" chứ không phải "\(< \)
Ta có : \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow\hept{\begin{cases}f\left(-3\right)=a.\left(-3\right)^2+b.\left(-3\right)+c=9a-3b+c\\f\left(4\right)=a.4^2+b.4+c=16a+4b+c\end{cases}}\)
\(\Rightarrow f\left(4\right)+f\left(-3\right)=\left(16a+4b+c\right)+\left(9a-3b+c\right)=25a+b+2c=0\)
\(\Rightarrow f\left(-3\right)+f\left(4\right)=0\)
\(\Rightarrow f\left(-3\right)=-f\left(4\right)\)
\(\Rightarrow f\left(-3\right).f\left(4\right)=-f\left(4\right).f\left(4\right)=-[f\left(4\right)]^2\le0\)\(\forall x\)
\(\Rightarrowđpcm\)
Vì f(0)=5 nên x*0+b*0+c=5
0+0+c=5 nên c=5
Vì f(1)=0 nên a*12+b*1+5=0
a+b+5=0
a+b=0-5
a+b=-5
Vì f(5)=0 nên a*52+b*5+5=0
5(5a+b+1)=0
5a+b+1=0/5=0
4a+a+b=0-1
4a+(-5)=-1
4a=-1-(-5)
4a=4
a=4/4
a=1
nên b=-5-1=-6
Vậy a=1;b=-6 và c=5
Ta co:
- f(0) = a.02+b.0+c = 0+0+c = c= 5
- f(1) = a.12+b.1+c = a+b+5 = 0 => a+b = -5
- f(5) = a.52+b.5+c = 25a + 5b + 5 = 0 => 25a+5b = -5
=> a+b = 25a+5b = -5
=> 25a-a + 5b-b = 0
=> 24a + 4b = 0
=> 24a = -4b
=> 24/-4 = b/a
=> b/a = -6
Tu \(\frac{b}{a}=-6=>\frac{b}{-6}=\frac{a}{1}=\frac{b+a}{-6+1}=-\frac{5}{-5}=1\)
=> a = 1 ; b=-6
Vay: a=1 ; b=-6 ; c =5
+f(0)= a.0 +b.0 + c =-3 => c = -3
+f(1) = a.12 +b.1-3 = 0 => a+b =3 (1)
+f(-1) = a(-1)+b(-1) -3 =-10 => a -b = -7 (2)
(1)(2) => a =(-7+3):2= -2
b =3-(-2) = 5