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a)\(f\left(1\right)=2.1^2+5.1-3=2+5-3=4\)
\(f\left(0\right)=0+0-3=-3\)
\(f\left(1,5\right)=2.\left(1,5\right)^2-5.1,5-3=4,5-7,5-3=-6\)
\(f\left(x\right)=x^2-5x+6\)
a) +) \(f\left(-\frac{1}{3}\right)=\left(-\frac{1}{3}\right)^2-5.\left(-\frac{1}{3}\right)+6=\frac{70}{9}\)
+) \(f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2-5.\frac{1}{2}+6=\frac{15}{4}\)
+) \(f\left(0\right)=0^2-5.0+6=6\)
+) \(f\left(1\right)=1^2-5.1+6=2\)
b) \(x^2-5x+6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
ok
1.\(f\left(x\right)=0\)
\(=>\left|3x-1\right|=0\)
\(=>3x-1=0\)
\(=>3x=1\)
\(=>x=\frac{1}{3}\)
\(f\left(x\right)=1\)
\(=>\left|3x-1\right|=1\)
\(=>\orbr{\begin{cases}3x-1=-1\\3x-1=1\end{cases}}\)
\(=>\orbr{\begin{cases}3x=-1+1=0\\3x=1+1=2\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
Vậy ...
Ta có hàm số : \(y=f\left(x\right)=ax-3\)
\(f\left(3\right)=9\)
\(=>ax-3=9\)
\(=>3a-3=9\)
\(=>3a=9+3=12\)
\(=>a=4\)
\(f\left(5\right)=11\)
\(=>ax-3=11\)
\(=>5a-3=11\)
\(=>5a=11+3=14\)
\(=>a=\frac{14}{5}\)
\(a,f\left(-1\right)=\left(-5\right)\left(-1\right)-3=5-3=2\\ f\left(\dfrac{2}{3}\right)=-5.\dfrac{2}{3}-3=\dfrac{-10}{3}-3=-\dfrac{19}{3}\)
\(b,y=-8\Rightarrow-8=-5x-3\Rightarrow-5=-5x\Rightarrow x=1\\ y=6\Rightarrow6=-5x-3\Rightarrow9=-5x\Rightarrow x=-\dfrac{9}{5}\)
a: f(-1)=5-3=2
f(2/3)=-10/3-3=-19/3
b: y=-8
=>-5x-3=-8
=>-5x=-5
hay x=1
y=6
=>-5x-3=6
=>-5x=9
hay x=-9/5
b,\(f\left(a\right)=2a-24\)
\(-f\left(-a\right)=-\left(2\left(-a\right)-24\right)=2a+24\)
Cộng cả 2 vế của BĐT \(-24< 24\), với \(2a\), ta có :
\(2a-24< 2a+24\)
Vậy \(f\left(a\right)< -f\left(-a\right)\)
a) f(1)= 2x1 - 8x3= 2-24=-22
f(-1)= -26; f(0,5)= -23; f(-0,5)= -25
b) f(a) > f(-a)