a.

\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)

b.

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

c.

\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)

a: \(\left(\frac{3x^2-18x-2}{1-2x}\right)^{\prime}=\frac{\left(3x^2-18x-2\right)^{\prime}\cdot\left(1-2x\right)-\left(3x^2-18x-2\right)\left(1-2x\right)^{\prime}}{\left(1-2x\right)^2}\)

\(=\frac{\left(6x-18\right)\left(1-2x\right)+2\left(3x^2-18x-2\right)}{\left(2x-1\right)^2}=\frac{6x-12x^2-18+36x+6x^2-36x-4}{\left(2x-1\right)^2}\)

\(=\frac{-6x^2+6x-22}{\left(2x-1\right)^2}\)

\(\left(\frac{2x-3}{x+4}\right)^{\prime}=\frac{\left(2x-3\right)^{\prime}\left(x+4\right)-\left(2x-3\right)\left(x+4\right)^{\prime}}{\left(x+4\right)^2}\)

\(=\frac{2\left(x+4\right)-\left(2x-3\right)}{\left(x+4\right)^2}=\frac{2x+8-2x+3}{\left(x+4\right)^2}=\frac{11}{\left(x+4\right)^2}\)

=>y'=\(\frac{-6x^2+6x-22}{\left(2x-1\right)^2}-\frac{11}{\left(x+4\right)^2}\)

\(=\frac{\left(-6x^2+6x-22\right)\left(x^2+8x+16\right)-11\left(2x-1\right)^2}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

\(=\frac{-6x^4-48x^3-96x^2+6x^3+48x^2+96x-22x^2-176x-352-11\left(4x^2-4x+1\right)}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

\(=\frac{-6x^4-42x^3-70x^2-80x-352-44x^2+44x-11}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

\(=\frac{-6x^4-42x^3-114x^2-36x-363}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

xét hàm số y=\(\sin\left(cos^2x\right)cos\left(sin^2x\right)\) =

\(\frac{sin\left(cos^2x+sin^2x\right)+sin\left(cos^2x-sin^2x\right)}{2}=\frac{sin1+sin\left(cós2x\right)}{2}\)

Chọn D.

Chọn A

a: \(0< =cos^23x< =1\)

=>\(9< =cos^23x+9< =10\)

=>9<=y<=10

\(y_{min}=9\) khi \(cos^23x=0\)

=>\(cos3x=0\)

=>3x=pi/2+kpi

=>x=pi/6+kpi/3

\(y_{max}=10\) khi \(cos^23x=0\)

=>\(sin^23x=0\)

=>3x=kpi

=>x=kpi/3

b: \(0< =sin^2x< =1\)

=>\(-3< =y< =-2\)

\(y_{min}=-3\) khi \(sin^2x=0\)

=>x=kpi

\(y_{max}=-2\) khi \(sin^2x=1\)

=>\(cos^2x=0\)

=>x=pi/2+kpi

c: \(0< =sin^25x< =1\)

=>12<=y<=13

y min=12 khi sin²5x=0

=>sin 5x=0

=>5x=kpi

=>x=kpi/5

y max=13 khi sin²5x=0

=>cos²5x=0

=>cos5x=0

=>5x=pi/2+kpi

=>x=pi/10+kpi/5

Đáp án là C.

1. Hàm số xác định `<=> 1-cosx \ne 0<=>cosx \ne 1<=>x \ne k2π`

Vì: `1+cosx >=0 forallx ; 1-cosx >=0 forall x`

2. Hàm số xác định `<=> sin^2x \ne cos^2x <=> (1-cos2x)/2 \ne (1+cos2x)/2`

`<=>cos2x \ne 0<=> 2x \ne π/2+kπ <=> x \ne π/4+kπ/2`

3. Hàm số xác định `<=> cos2x \ne 0<=> x \ne π/4+kπ/2 (k \in ZZ)`.

Bạn cho mình hỏi tại sao x khác k2\(\pi\) là lý thuyết ở đoạn nào thế ạ?