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Bài 1:
\(a)f\left(x\right)=10x\)
\(\Leftrightarrow f\left(0\right)=10.0=0\)
\(\Leftrightarrow f\left(-1\right)=10\left(-1\right)=-10\)
\(\Leftrightarrow f\left(\frac{1}{2}\right)=\frac{10}{2}=5\)
\(b)\)Vì \(f\left(x\right)=10x\)
Nên: \(f\left(a+b\right)=10\left(a+b\right)\)
Và: \(f\left(a\right)+f\left(b\right)=10a+10b=10\left(a+b\right)\)
Do đó:
\(f\left(a+b\right)=f\left(a\right)+f\left(b\right)\left(đpcm\right)\)
\(c)\)Vì \(\hept{\begin{cases}f\left(x\right)=10x\\f\left(x\right)=x^2\end{cases}\Leftrightarrow x^2=10x}\)
\(\Leftrightarrow x^2-10x=0\)
\(\Leftrightarrow x\left(x-10\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=10\end{cases}}}\)
Vậy với \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)thì \(f\left(x\right)=x^2\)
\(a.\)
Theo đề , ta có : \(y=f\left(x\right)=4x^2-5\)
\(\Rightarrow\)
\(f\left(3\right)=4.\left(3\right)^2-5=31\)
\(f\left(-\frac{1}{2}\right)=4.\left(-\frac{1}{2}\right)^2-5=-4\)
\(b.\)
Ta có : \(f\left(x\right)=-1\)
\(\Rightarrow4x^2-5=-1\)
\(\Rightarrow4x^2=-1+5=4\)
\(\Rightarrow x^2=4:4=1\)
\(\Rightarrow x=\sqrt{1}=1\)
\(c.\)
Ta có :
\(f\left(x\right)=4x^2-5\)
\(\Rightarrow f\left(x\right)=4.\left(x\right)^2-5\) \(\left(1\right)\)
\(f\left(-x\right)=4.\left(-x\right)^2-5=4.\left(x\right)^2-5\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow f\left(x\right)=f\left(-x\right)\)
Lười làm quá!
a) f ( 1/2 ) = 4 . ( 1/2 )2 - 7 = 4 . 1/4 - 7 = 1 - 7 = - 6
f ( 3 ) = 4 . 32 - 7 = 4 . 9 - 7 = 36 - 7 = 29
f 0 ) = 4 . 02 - 7 = 4 . 0 - 7 = 0 - 7 = - 7
f ( - 2 ) = 4 . ( - 2 )2 - 7 = 4 . 8 - 7 = 32 - 7 = 25
b) f ( x ) = 93
4 . x2 - 7 = 93
=> 4 . x2 = 93 + 7
=> 4 . x2 = 100
=> x2 = 100 : 4
=> x2 = 25
=> x2 = 52
=. x = 5 hoặc x = - 5
Vậy ...
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) Cho hàm số y = f(x) = -3x\(^2\)+1
f\(\left(\frac{-1}{2}\right)\) = -3.\(\left(\frac{-1}{2}\right)\)\(^2\)+1 = -3.\(\frac{1}{4}\)+1 = \(\frac{-3}{4}\)+\(\frac{4}{4}\) = \(\frac{1}{4}\)
f\(\left(\frac{1}{3}\right)\) = -3.\(\left(\frac{1}{3}\right)^2\)+1 = -3.\(\frac{1}{9}\)+1 = \(\frac{-1}{3}+\frac{3}{3}=\frac{2}{3}\)
f\(\left(0\right)=-3.0^2+1=-3.0+1=0+1=1\)
f(-1) = \(-3.\left(-1\right)^2+1=-3.1+1=-3+1=-2\)
b) Cho hàm số y = f(x) = 2-x\(^2\)
f(2) = \(2-2^2=2-4=-2\)
\(f\left(1\right)=2-1^2=2-1=1\)
f(0) = \(2-0^2=2-0=2\)
f(-2) = \(2-\left(-2\right)^2=2-4=-2\)
a/ ta có: f(0)=9*02-2=-2
f(-1/3)=9*(-1/3)2-2=-1
f(\(3\sqrt{2}\)