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

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

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

\(y'\left(0\right)=\dfrac{1}{\left(0+1\right)^2}=1\)

a) Ta có: 

\(f\left(-2\right)=\left|3\cdot-2-1\right|=\left|-6-1\right|=\left|-7\right|=7\) 

\(f\left(2\right)=\left|3\cdot2-1\right|=\left|6-1\right|=5\)

\(f\left(-\dfrac{1}{4}\right)=\left|3\cdot-\dfrac{1}{4}-1\right|=\left|-\dfrac{3}{4}-1\right|=\left|-\dfrac{7}{4}\right|=\dfrac{7}{4}\) 

b) Ta có: 

\(f\left(x\right)=10\)

\(\Rightarrow\left|3x-1\right|=10\)

Với \(x\ge\dfrac{1}{3}\Rightarrow3x-1=10\)

\(\Rightarrow3x=11\Rightarrow x=\dfrac{11}{3}\left(tm\right)\)

Với \(x< \dfrac{1}{3}\Rightarrow3x-1=-10\)

\(\Rightarrow3x=-9\Rightarrow x=-3\left(tm\right)\) 

_______

\(f\left(x\right)=-3\)

\(\Rightarrow\left|3x-1\right|=-3\)

Mà: \(\left|3x-1\right|\ge0\forall x\) và \(-3< 0\)

\(\Rightarrow\left|3x-1\right|=-3\) (vô lý)

\(\Rightarrow\) không có x thỏa mãn 

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 \(x=\sqrt{2}-1\)

a)  Cho hàm số y = f(x) = -2x + 3.

Ta có: f(-2)= -2.(-2)+3

                 = 4+3=7

Ta có: f(0)= -2.0+3

                = 0+3=3

Ta có: f(

Lời giải:

a.

$f(-2)=(-2)(-2)+3=7$

$f(0)=(-2).0+3=3$

$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$

b.

$f(x)=-2x+3=5$

$\Rightarrow -2x=2$

$\Rightarrow x=-1$

$f(x)=-2x+3=1$

$\Rightarrow -2x=1-3=-2$

$\Rightarrow x=1$

Tham khảo:

\(a,F=\dfrac{x^2+x+4x^2+2-x^2+3x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x}{x-1}\\ b,\left|x+2\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1-2=-1\left(ktm\right)\\x=-1-2=-3\end{matrix}\right.\Leftrightarrow x=-3\\ \Leftrightarrow F=\dfrac{-12}{-4}=3\\ c,K=F\left(x-1\right)-x^2-2021=4x-x^2-2021\\ K=-\left(x^2-4x+4\right)-2017=-\left(x-2\right)^2-2017\le-2017\\ K_{max}=-2017\Leftrightarrow x=2\left(tm\right)\)

Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)

A. √3+1/2         B. √3−1/2          C. 1−√3/2             D. 0

B

\(f'\left(x\right)-f\left(x\right)=2cosx\)

\(\Leftrightarrow e^{-x}.f'\left(x\right)-e^{-x}.f\left(x\right)=2e^{-x}cosx\)

\(\Rightarrow\left[e^{-x}.f\left(x\right)\right]'=2e^{-x}.cosx\)

Lấy nguyên hàm 2 vế:

\(\Rightarrow e^{-x}.f\left(x\right)=\int2e^{-x}cosxdx=e^{-x}\left(sinx-cosx\right)+C\)

Thay \(x=\dfrac{\pi}{2}\Rightarrow e^{-\dfrac{\pi}{2}}.1=e^{-\dfrac{\pi}{2}}+C\Rightarrow C=0\)

\(\Rightarrow f\left(x\right)=sinx-cosx\)

\(\Rightarrow f\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}-1}{2}\)