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\(f\left(2\right)=-2.2^2+3=-2.4+3=-5\)
\(f\left(1\right)=-2.1^2+3=-2+3=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ta có f (x ) = 2 x ^2 -1
=) f (1) = 2 *1 ^2 -1
=) f (1) =1
; ta có f (-1/2) = 2 * -1/2 ^ 2=1 = -3/2
theo mình thì chắc k phải tìm x mà phải giả sử x bằng 1 số bất kì nào đó rồi tìm hàm số .b ) cho x= 2
=) 3= f (2) = 2* 2^2 -1=7
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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 \(f\left(x\right)=x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : \(f\left(0\right)=2.0^2-10=-10\)
\(f\left(1\right)=2.1^2-10=-8\)
\(f\left(-1\frac{1}{2}\right)=f\left(\frac{-3}{2}\right)=2.\left(\frac{-3}{2}\right)^2-10=2.\frac{9}{4}-10=\frac{9}{2}-10=\frac{-11}{2}\)
b)Vì \(f\left(x\right)=2\)
\(\Rightarrow2x^2-10=-2\)
\(\Rightarrow2x^2=8\)
\(\Rightarrow x^2=4\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy \(x=2\)hoặc \(x=-2\)
a, \(f\left(0\right)=2.0^2-10=-10\)
\(f\left(1\right)=2.1^2-10=2-10=-8\)
Ta co \(-1\frac{1}{2}=-\frac{3}{2}\)
\(f\left(-\frac{3}{2}\right)=2.\left(-\frac{3}{2}\right)^2-10=2.\frac{9}{4}-10=\frac{18}{4}-\frac{40}{4}=-\frac{22}{4}=-\frac{11}{2}\)
b, Ta co : \(f\left(x\right)=-2\)hay \(2x^2-10=-2\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :+)\(f\left(x\right)=6-4x\)
\(\implies\)\(f\left(1\right)=6-4=2\)
Ta có :+) \(f\left(x\right)=6-4x=\frac{1}{2}\)
\(\implies\) \(x=\frac{11}{8}\)
Ta có :+) \(g\left(x\right)=2x^2-3x\)
\(\implies\) \(g\left(-2\right)=2.\left(-2\right)^2+3.2\)
\(\implies\) \(g\left(-2\right)=4.2+3.2\)
\(\implies\) \(g\left(-2\right)=14\)
Ta có :
f(2) = -22 + 2 . 2 = -4 + 4 = 0
f(-3) = -32 + 2 . ( -3 ) = -9 + ( -6 ) = -15
f(1/2 ) = \(-\left(\frac{1}{2}\right)^2+2.\frac{1}{2}=-\frac{1}{4}+1=\frac{3}{4}\)