\(y'=\dfrac{\left[tan^3\left(2x+1\right)\right]'}{2\sqrt{tan^3\left(2x+1\right)}}\)

\(=\dfrac{3\cdot tan^2\left(2x+1\right)\cdot\left(tan\left(2x+1\right)\right)'}{2\sqrt{tan^3\left(2x+1\right)}}\)

\(=\dfrac{3}{2}\cdot\sqrt{tan\left(2x+1\right)}\cdot\dfrac{1}{cos^2\left(2x+1\right)}\)

y=2

=>x^2-5x+6=2

=>x^2-5x+4=0

=>x=1 hoặc x=4

y'=2x-5

Khi x=1 thì y'=2-5=-3

Khi x=4 thì y'=2*4-5=8-5=3

y-y0=f'(x-x0)

Khi x=1 thì y-2=-3(x-1)

=>y=-3x+3+2=-3x+5

Khi x=4 thì y-2=3(x-4)

=>y=3x-12+2=3x-10

1:

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

\(=\dfrac{2x+4-2x+1}{\left(x+2\right)^2}=\dfrac{5}{\left(x+2\right)^2}\)

2:

\(\left(2x^2-3x+1\right)'=2\cdot2x-3=4x-3\)

(1-x^2)'=1-2x

\(y'=\left(4x-3\right)\cdot\left(1-x^2\right)+\left(2x^2-3x+1\right)\left(1-2x\right)\)

\(=4x-4x^3-3+3x^2+2x^2-4x^3-3x+6x^2+1-2x\)

\(=-8x^3+11x^2-x-2\)

5:

\(\sqrt{2x^2+1}'=\dfrac{\left(2x^2+1\right)'}{\sqrt{2x^2+1}}=\dfrac{4x}{\sqrt{2x^2+1}}\)

\(y'=x'+\sqrt{2x^2+1}'=1+\dfrac{4x}{\sqrt{2x^2+1}}\)

(SD;(ABCD))=(DS;DA)=góc SDA

tan SDA=SA/AD=3/2

=>góc SDA=56 độ

1: BC vuông góc AB

BC vuông góc SA

=>BC vuông góc (SAB)

=>(SAB) vuông góc (SBC)

SỐ cách chọn ngẫu nhiên 4 viên bi là:

\(C^4_{12}=495\left(cách\right)\)

\(\overline{abc}\)

a có 5 cách

b có 5 cách

c có 4 cách

=>Có 5*5*4=100 cách