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 \(2x^2-10=-2\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

(1)

a) x=\(\dfrac{-1}{12}-\dfrac{2}{3}\)=\(\dfrac{-3}{4}\)

b) 2x+1=3 => 2x=3-1=2 => x=1

(2)

f(2)=2.2²+4=12

f(-1)=2.(-1)²+4=6

(1)

a) \(x+\dfrac{2}{3}=-\dfrac{1}{12}\\ \Rightarrow x=-\dfrac{1}{12}-\dfrac{2}{3}\\ \Rightarrow x=\dfrac{-1}{12}-\dfrac{8}{12}\\ \Rightarrow x=-\dfrac{9}{12}=-\dfrac{3}{4}\)

Vậy \(x=-\dfrac{3}{4}\)

b) \(\left(2x+1\right)^2=9\\ \Rightarrow\left(2x+1\right)^2=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-2;1\right\}\)

(2)

\(y=f\left(x\right)=2x^2+4\\ f\left(2\right)=2\cdot2^2+4=8+4=12\\ f\left(-1\right)=2\cdot\left(-1\right)^2+4=2+4=6\)

Vậy \(f\left(2\right)=12\\ f\left(-1\right)=6\)

\(\left[{}\begin{matrix}f\left(-1\right)=-1^2+2\cdot-1-1=-2\\f\left(0\right)=0^2+2\cdot0-1=-1\\f\left(1\right)=1^2+2\cdot1-1=2\end{matrix}\right.\)

Gửi bạn

a: f(-1/2)=17/4

f(5)=29

\(a,f\left(-\dfrac{1}{2}\right)=\dfrac{1}{4}+4=\dfrac{17}{4}\\ f\left(5\right)=25+4=29\\ b,f\left(x\right)=10=x^2+4\Leftrightarrow x^2=6\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\)

\(\text{1)}\)

\(\text{Thay }x=-2,\text{ ta có: }f\left(-2\right)-5f\left(-2\right)=\left(-2\right)^2\Rightarrow f\left(-2\right)=-1\)

\(\Rightarrow f\left(x\right)=x^2+5f\left(-2\right)=x^2-5\)

\(f\left(3\right)=3^2-5\)

\(\text{2)}\)

\(\text{Thay }x=1,\text{ ta có: }f\left(1\right)+f\left(1\right)+f\left(1\right)=6\Rightarrow f\left(1\right)=2\)

\(\text{Thay }x=-1,\text{ ta có: }f\left(-1\right)+f\left(-1\right)+2=6\Rightarrow f\left(-1\right)=2\)

\(\text{3)}\)

\(\text{Thay }x=2,\text{ ta có: }f\left(2\right)+3f\left(\frac{1}{2}\right)=2^2\text{ (1)}\)

\(\text{Thay }x=\frac{1}{2},\text{ ta có: }f\left(\frac{1}{2}\right)+3f\left(2\right)=\left(\frac{1}{2}\right)^2\text{ (2)}\)

\(\text{(1) - 3}\times\text{(2) }\Rightarrow f\left(2\right)+3f\left(\frac{1}{2}\right)-3f\left(\frac{1}{2}\right)-9f\left(2\right)=4-\frac{1}{4}\)

\(\Rightarrow-8f\left(2\right)=\frac{15}{4}\Rightarrow f\left(2\right)=-\frac{15}{32}\)

sai 1 chút chỗ cÂU 3

nhân vs 3 thì phải là 1/12

thay x bằng ? mik cũng ko bit làm lên vào đây tham khảo hihihihi

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 : \(y=f\left(x\right)=ax-3\)

\(f\left(3\right)=9\)

\(=>ax-3=9\)

\(=>3a-3=9\)

\(=>3a=9+3=12\)

\(=>a=4\)

\(f\left(5\right)=11\)

\(=>ax-3=11\)

\(=>5a-3=11\)

\(=>5a=11+3=14\)

\(=>a=\frac{14}{5}\)

Ta có hàm số \(y=f\left(x\right)=ax+b\)

\(f\left(0\right)=1\)

\(=>ax+b=1\)

\(=>a.0+b=1\)

\(=>b=1\)

\(f\left(-1\right)=2\)

\(=>ax+b=2\)

\(=>-a+1=2\)

\(=>-a=2-1=1\)

\(=>a=-1\)

vậy ...

a: TXĐ: D=R

b: \(f\left(-1\right)=\dfrac{2}{-1-1}=\dfrac{2}{-2}=-1\)

\(f\left(0\right)=\sqrt{0+1}=1\)

\(f\left(1\right)=\sqrt{1+1}=\sqrt{2}\)

\(f\left(2\right)=\sqrt{3}\)