Áp dụng BĐT \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\):

\(y^2=\left(\sqrt{sinx}+\sqrt{1-sinx}\right)^2\le sinx+1-sinx=1\)

\(\Rightarrow-1\le y\le1\)

\(\Rightarrow M^4-m^4=0\)

1.

Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng

2.a

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)

\(\Leftrightarrow1+tan^2x=4tanx+6\)

\(\Leftrightarrow tan^2x-4tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)

2b.

Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)

\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin^3t=sint+cost\)

\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)

\(\Leftrightarrow sint.cos^2t+cost=0\)

\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

2c.

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)

\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)

\(\Leftrightarrow2sinx.cos2x+cos2x=1\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)

\(\Leftrightarrow1-2sin^2x-sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)

Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)

Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

ĐKXĐ: \(x\ge1\)

\(3\sqrt[]{x-1}+m\sqrt[]{x+1}=2\sqrt[4]{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow3\sqrt[]{\dfrac{x-1}{x+1}}+m=2\sqrt[4]{\dfrac{x-1}{x+1}}\)

Đặt \(\sqrt[4]{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)

\(\Rightarrow3t^2+m=2t\Leftrightarrow-3t^2+2t=m\)

Xét \(f\left(t\right)=-3t^2+2t\) trên \([0;1)\)

\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)

\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)

\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)

\(\Rightarrow-1< m\le\dfrac{1}{3}\)

Chọn C

Coi như tất cả các biểu thức cần tính đạo hàm đều xác định.

1.

\(y'=2sin\sqrt{4x+3}.\left(sin\sqrt{4x+3}\right)'=2sin\sqrt{4x+3}.cos\sqrt{4x+3}.\left(\sqrt{4x+3}\right)'\)

\(=sin\left(2\sqrt{4x+3}\right).\dfrac{4}{2\sqrt{4x+3}}=\dfrac{2sin\left(2\sqrt{4x+3}\right)}{\sqrt{4x+3}}\)

2.

\(y'=3x^3+\dfrac{17}{x\sqrt{x}}\)

3.

\(y'=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\left(\dfrac{sin4x}{cos\left(x^2+2\right)}\right)'\)

\(=\dfrac{1}{2\sqrt{\dfrac{sin4x}{cos\left(x^2+2\right)}}}.\dfrac{4cos4x.cos\left(x^2+2\right)+2x.sin4x.sin\left(x^2+2\right)}{cos^2\left(x^2+2\right)}\)

4.

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

\(=-\dfrac{2sin\left(6-x\right).\left[sin\left(6-x\right)\right]'+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}=-\dfrac{-2sin\left(6-x\right).cos\left(6-x\right)+4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

\(=\dfrac{sin\left(12-2x\right)-4}{2\sqrt{\left[sin^2\left(6-x\right)+4x\right]^3}}\)

5.

\(y'=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).\left[sin\left(\dfrac{2x-1}{4-x}\right)\right]'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+2x.sin\left(\dfrac{2x-1}{4-x}\right).cos\left(\dfrac{2x-1}{4-x}\right).\left(\dfrac{2x-1}{4-x}\right)'\)

\(=sin^2\left(\dfrac{2x-1}{4-x}\right)+x.sin\left(\dfrac{4x-2}{4-x}\right).\dfrac{7}{\left(4-x\right)^2}\)

8.

\(y=tan^33x-\left(sin2x+cos3x\right)^5\)

\(\Rightarrow y'=3tan^23x.\left(tan3x\right)'-5\left(sin2x+cos3x\right)^4.\left(sin2x+cos3x\right)'\)

\(=\dfrac{9.tan^23x}{cos^23x}-5\left(sin2x+cos3x\right)^4.\left(2cos2x-3sin3x\right)\)

9.

\(y'=6cot^55x.\left(cot5x\right)'-4cos^33x.\left(cos3x\right)'+3cos3x\)

\(=-\dfrac{30.cot^55x}{sin^25x}+12cos^33x.sin3x+3cos3x\)

a) \(y=\sqrt{1-sin\left(x^2\right)}-1\)  đạt giá trị lớn nhất là 1 , giá trị nhỏ nhất là - 1 ( để ý rằng u = x + \(\frac{\pi}{3}\) lấy mọi giá trị thực tùy ý khi x thay đổi ) , nên hàm số y = 2cos \(\left(x+\frac{\pi}{3}\right)\) + 3 đạt giá trị lớn nhất là y = 2 . 1 + 3 = 5 , giá trị nhỏ nhất là y = 2 . ( - 1 ) + 3 = 1

b) Hàm số y = 4sin |x| = đạt giá trị lớn nhất là 4 ( khi sin | x | = 1 tức là | x | = \(\frac{\pi}{2}\) + 2k\(\pi\) , k nguyên không âm ) , đạt giá trị nhỏ nhất - 4 ( khi sin | x | = \(-\frac{\pi}{2}+2k\pi\) , k nguyên dương )