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a) \(f\left(\frac{3}{2}\right)=2.\frac{3}{2}=3\)
b) \(f\left(a\right)+f\left(-a\right)=2a+\left(-2a\right)=0\left(đpcm\right)\)
\(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)
\(f\left(3\right)=a.3^2+b.3+c=9a+3b+c\)
\(f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\Rightarrow f\left(-2\right).f\left(3\right)\le0\)
a) \(f\left(3\right)=4\times3^2-5=31\)
\(f\left(-\frac{1}{2}\right)=4\times\left(-\frac{1}{2}\right)^2-5=-4\)
b) để f(x)=-1
<=>\(4x^2-5=-1\)
<=>\(4x^2=4\)
<=>\(x^2=1\)
<=>\(x=\orbr{\begin{cases}1\\-1\end{cases}}\)
Cho hàm số y = f(x) = 4x^2 +4y=f(x)=4x2+4. Tính f(-2)f(−2) ; f(2)f(2) ; f(4)f(4).
Đáp số:
f(-2) =f(−2)=
f(2) =f(2)=
f(4) =f(4)=
\(a.\)
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 \(\left(2\right)\Rightarrow f\left(x\right)=f\left(-x\right)\)
Nguyễn Việt Lâm Trần Trung Nguyên tran nguyen bao quan Shurima Azir Nguyễn Thanh Hằng Mysterious Person Phùng Khánh Linh Aki Tsuki
a) theo tính chất ta có: f(0+0)= f(0)+f(0)
=> f(0)=f(0)+f(0)
=> f(0)-f(0)=f(0)+f(0)-f(0)
=> 0=f(0)
hay f(0)=0
b) f(0)=f(-x+x)=f(-x)+f(x)
=>0=f(-x)+f(x)
=> f(-x)=0-f(x)=-f(x)
c) \(f\left(x_1-x_2\right)=f\left(x_1+\left(-x_2\right)\right)=f\left(x_1\right)+f\left(-x_2\right)=f\left(x_1\right)-f\left(x_2\right)\)
a) f(-2)=5 – 2. (-2) = 5 + 4 = 9;
f(-1) = 5 – 2.(-1) = 5 + 2 = 7;
f(0) = 5 – 2.0 = 5;
f(3) = 5 – 2.3 = 5 – 6 = -1.
b)\(y=5-2x\Rightarrow x=\dfrac{5y}{2}\)
\(y=5\Rightarrow x=\dfrac{5-5}{2}=0\)
\(y=3\Rightarrow x=\dfrac{5-3}{2}=1\)
\(y=-1\Rightarrow x=\dfrac{5-\left(-1\right)}{2}=\dfrac{5+1}{2}=3\)
Ta có : \(f\left(a\right)=5\cdot a^2+1\)
và \(f\left(-a\right)=5\cdot\left(-a\right)^2+1\)
Dễ thấy \(5\cdot a^2=5\cdot\left(-a\right)^2\)
Do đó : \(5\cdot a^2+1=5\cdot\left(-a\right)^2+1\left(đpcm\right)\)
f(-a)=5(-a)2+1=5a+1
f(a)=5a+1
Vậy f(-a)=f(a)
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: f(3/2)=2*3/2=3
b: f(a)=2a
f(-a)=-2a
=>f(a)+f(-a)=0