a,

Khi f(3)

=> 5 . 3² - 1

= 5 . 9 - 1

= 45 - 1

= 44

Khi f(-2)

=> 5 . ( -2 )² - 1

= 5 . 4 - 1

= 20 - 1

= 19

b,

Khi f(x) = 79

=> 5x² - 1 = 79

5x² = 79 + 1

5x² = 80

=> x² = 80 : 5

x² = 16

x² = 4²

=> x = 4

a)\(f\left(3\right)=5\cdot3^2-1=5\cdot9-1=45-1=44\)

\(f\left(-2\right)=5\cdot\left(-2\right)^2-1=5\cdot4-1=20-1=19\)

b)\(f\left(x\right)=79\Leftrightarrow5x^2-1=79\)

\(\Leftrightarrow5x^2=80\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x=\pm4\)

f(1)=m^2+2m+1

h(-1)=m^2-2m

f(1)=h(-1)=> m^2+2m+1=m^2-2m=> m=-1/4

Câu 1: 

a) 

b) Ta có: f(x)=8

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)

Ta có: \(f\left(x\right)=6-4\sqrt{2}\)

\(\Leftrightarrow2x^2=6-4\sqrt{2}\)

\(\Leftrightarrow x^2=3-2\sqrt{2}\)

\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)

hay \(x=\sqrt{2}-1\)

Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)

1/ \(y'=\dfrac{\left(\sqrt{x+1}\right)'x-x'\sqrt{x+1}}{x^2}=\dfrac{\dfrac{x}{2\sqrt{x+1}}-\sqrt{x+1}}{x^2}=\dfrac{-x-2}{2x^2\sqrt{x+1}}\)

2/ \(y'=\dfrac{1-x^2-\left(1-x^2\right)'x}{\left(1-x^2\right)^2}=\dfrac{1+x^2}{\left(1-x^2\right)^2}\)

3/ \(y'=\dfrac{-\left(x-\sqrt{x+1}\right)'}{\left(x-\sqrt{x+1}\right)^2}=\dfrac{-1+\dfrac{1}{2\sqrt{x+1}}}{\left(x-\sqrt{x+1}\right)^2}\)

4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)

\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)

5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)

Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học

ok cảm ơn bạn nhìu

\(a,f\left(0\right)=-2\\ f\left(-4\right)=3.16-2=46\\ b,y=25\Leftrightarrow3x^2-2=25\Leftrightarrow x^2=9\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

hảo hán !

cảm ơn bạn rất nhiều ạ