a: ĐKXĐ; 1-sin x>=0

=>sin x<=1(luôn đúng)

b: ĐKXĐ: 1-cosx>=0

=>cosx<=1(luôn đúng)

c: ĐKXĐ: 1-cos²x>=0

=>cos²x<=1

=>-1<=cosx<=1(luôn đúng)

Bạn chú ý gõ đề bằng công thức toán (hộp biểu tượng $\sum$) trên thanh công cụ. Nhìn đề rối mắt thế này thật tình không ai muốn đọc chứ đừng nói đến giúp =)))

1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)

⇔ \(\dfrac{\left(\sqrt{3}-1\right)}{2\sqrt{2}}sinx+\dfrac{\left(\sqrt{3}+1\right)}{2\sqrt{2}}cosx=\dfrac{1-\sqrt{3}}{2\sqrt{2}}\)

⇔ sinx . si

b)  \(3\left(1-2x\right)^{20}\left(3x-2\right)^{10}\left(-14\left(3x-2\right)+11\left(1-2x\right)\right)\)

1.

ĐKXĐ: \(\frac{1+x}{1-x}\ge0\Leftrightarrow-1\le x< 1\)

2.

\(cosx-cos3x\ne0\)

\(\Leftrightarrow cos3x\ne cosx\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)

3.

a. \(0\le\left|sinx\right|\le1\Rightarrow1\le y\le3\)

\(y_{min}=1\) khi \(\left|sinx\right|=1\)

\(y_{max}=3\) khi \(sinx=0\)

b. \(y=cosx+cos\left(x-\frac{\pi}{3}\right)=2cos\left(x-\frac{\pi}{6}\right).cos\frac{\pi}{6}=\sqrt{3}cos\left(x-\frac{\pi}{6}\right)\)

\(-1\le cos\left(x-\frac{\pi}{6}\right)\le1\Rightarrow-\sqrt{3}\le y\le\sqrt{3}\)

c. \(y=cos^22x+2cos2x+1-1=\left(cos2x+1\right)^2-1\ge-1\)

\(y_{min}=-1\) khi \(cos2x=-1\)

\(cos2x\le1\Leftrightarrow\left\{{}\begin{matrix}cos^22x\le1\\2cos2x\le2\end{matrix}\right.\) \(\Rightarrow y\le3\)

\(y_{max}=3\) khi \(cos2x=1\)

d. \(5-2cos^2x.sin^2x=5-\frac{1}{2}\left(2sinx.cosx\right)^2=5-\frac{1}{2}sin^22x\)

\(0\le sin^22x\le1\Rightarrow\frac{9}{2}\le5-\frac{1}{2}sin^22x\le5\)

\(\Rightarrow\sqrt{\frac{9}{2}}\le y\le\sqrt{5}\)

a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)

\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)

\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)

Đến đây là dễ r nha bn :3

a: Ta có: \(y=\left(x+1\right)\left(\sqrt{x}-1\right)\)

=>\(y^{\prime}=\left(x+1\right)^{\prime}\cdot\left(\sqrt{x}-1\right)+\left(x+1\right)\cdot\left(\sqrt{x}-1\right)^{\prime}=\left(\sqrt{x}-1\right)+\left(x+1\right)\cdot\left(\frac{1}{2\sqrt{x}}\right)=\frac{x+1}{2\sqrt{x}}+\sqrt{x}-1\)

\(=\frac{x+1+2x-2\sqrt{x}}{2\sqrt{x}}=\frac{3x-2\sqrt{x}+1}{2\sqrt{x}}\)

b: ta có: \(y=\left(x^2-3\right)\left(x^3+3x^2-5\right)\)

=>\(y^{\prime}=\left(x^2-3\right)^{\prime}\left(x^3+3x^2-5\right)+\left(x^2-3\right)\left(x^3+3x^2-5\right)^{\prime}\)

=>\(y^{\prime}=2x\left(x^3+3x^2-5\right)+\left(x^2-3\right)\left(3x^2+6x\right)\)

\(=2x^4+6x^3-10x+3x^4+6x^3-9x^2-18x=5x^4+12x^3-9x^2-28x\)