Đề sai. Bạn xem lại đề nhé.

\(sinx.cos^3x-sin^3x.cosx\)

\(=sinx.cosx\left(cos^2x-sin^2x\right)\)

\(=\dfrac{1}{2}sin2x\left(cos^2x-sin^2x\right)\)

\(=\dfrac{1}{2}sin2x.cos2x\)

\(=\dfrac{sin4x}{4}\)

đề đúng không vậy

e/

Đề câu này chắc chắn đúng chứ bạn?

f/

\(sin^4x+cos^4x=\frac{3}{4}\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)

\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)

\(\Leftrightarrow1-2sin^22x=0\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

c/

\(y=sin\left(4x-\frac{\pi}{3}\right)+sin\left(\frac{\pi}{3}\right)+5\)

\(=sin\left(4x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}+5\)

Do \(-1\le sin\left(4x-\frac{\pi}{3}\right)\le1\)

\(\Rightarrow4+\frac{\sqrt{3}}{2}\le y\le6+\frac{\sqrt{3}}{2}\)

d/

\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin2x+5\)

\(y=6-3sin^2x.cos^2x+3sin2x\)

\(y=-\frac{3}{4}sin^22x+3sin2x+6\)

\(y=\frac{3}{4}\left(sin2x+1\right)\left(5-sin2x\right)+\frac{9}{4}\ge\frac{9}{4}\)

\(y_{min}=\frac{9}{4}\) khi \(sin2x=-1\)

\(y=\frac{3}{4}\left(sin2x-1\right)\left(3-sin2x\right)+\frac{33}{4}\le\frac{33}{4}\)

\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)

Bài 3:

1: \(\sqrt3\cdot\sin x+cosx=1\)

=>\(\sin x\cdot\frac{\sqrt3}{2}+cosx\cdot\frac12=\frac12\)

=>\(\sin\left(x+\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}x+\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ x+\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=k2\pi\\ x=\frac56\pi-\frac{\pi}{6}+k2\pi=\frac23\pi+k2\pi\end{array}\right.\)

2: \(\sin4x-\sqrt3\cdot cos4x=\sqrt2\)

=>\(\sin4x\cdot\frac12-cos4x\cdot\frac{\sqrt3}{2}=\frac{\sqrt2}{2}\)

=>\(\sin\left(4x-\frac{\pi}{3}\right)=\sin\left(\frac{\pi}{4}\right)\)

=>\(\left[\begin{array}{l}4x-\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\ 4x-\frac{\pi}{3}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=\frac{\pi}{3}+\frac{\pi}{4}+k2\pi=\frac{7}{12}\pi+k2\pi\\ 4x=\frac34\pi+\frac{\pi}{3}+k2\pi=\frac{13}{12}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{7}{48}\pi+\frac{k\pi}{2}\\ x=\frac{13}{48}\pi+\frac{k\pi}{2}\end{array}\right.\)

3: \(\sqrt3\cdot cos3x+\sin3x=\sqrt2\)

=>\(\sin3x\cdot\frac12+cos3x\cdot\frac{\sqrt3}{2}=\frac{\sqrt2}{2}\)

=>\(\sin\left(3x+\frac{\pi}{3}\right)=\sin\left(\frac{\pi}{4}\right)\)

=>\(\left[\begin{array}{l}3x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\ 3x+\frac{\pi}{3}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=\frac{\pi}{4}-\frac{\pi}{3}+k2\pi=-\frac{\pi}{12}+k2\pi\\ 3x=\frac34\pi-\frac{\pi}{3}+k2\pi=\frac{5}{12}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\ x=\frac{5}{36}\pi+\frac{k2\pi}{3}\end{array}\right.\)

4: \(cos7x\cdot cos5x-\sqrt3\cdot\sin2x=1-\sin7x\cdot\sin5x\)

=>\(cos7x\cdot cos5x+\sin7x\cdot\sin5x-\sqrt3\cdot\sin2x=1\)

=>cos(7x-5x)\(-\sqrt3\cdot\sin2x=1\)

=>\(\frac12\cdot cos2x-\frac{\sqrt3}{2}\cdot\sin2x=\frac12\)
=>\(\sin\left(\frac{\pi}{6}-2x\right)=\frac12\)

=>\(\left[\begin{array}{l}\frac{\pi}{6}-2x=\frac{\pi}{6}+k2\pi\\ \frac{\pi}{6}-2x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-2x=k2\pi\\ -2x=\frac46\pi+k2\pi=\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-k\pi\\ x=-\frac13\pi-k\pi\end{array}\right.\)

a) Ta có: \(3x+2\sqrt{3x}+4=\left(\sqrt{3x}+1\right)^2+3>0;1+\sqrt{3x}>0,\forall x\ge0\), nên đk để A có nghĩa là

\(\left(\sqrt{3x}\right)^3-8-\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)\ne0;x\ge0\Leftrightarrow\sqrt{3x}\ne2\Leftrightarrow0\le x\ne\frac{4}{3}\)

A=\(\left(\frac{6x+4}{\left(\sqrt{3x}\right)^3-2^3}-\frac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\frac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)

\(=\left(\frac{6x+4-\left(\sqrt{3x}-2\right)\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-\sqrt{3x}+1-\sqrt{3x}\right)\)

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

\(=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\left(0\le x\ne\frac{4}{3}\right)\)

b) \(A=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}=\frac{\left(\sqrt{3x}-2\right)^2+2\left(\sqrt{3x}-2\right)+1}{\sqrt{3x}-2}=\sqrt{3x}+\frac{1}{\sqrt{3x}-2}\)

Với \(x\ge0\), để A là số nguyên thì \(\sqrt{3x}-2=\pm1\Leftrightarrow\orbr{\begin{cases}\sqrt{3x}=3\\\sqrt{3x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=9\\3x=1\end{cases}\Leftrightarrow}x=3}\)  (vì \(x\in Z;x\ge0\))

Khi đó A=4