\(\sqrt{x}+\dfrac{1}{\sqrt{x}}+\dfrac{x^{10}}{10}=U+V+T\)

\(\left\{{}\begin{matrix}U^2=x;\\V^2=\dfrac{1}{x}\\Y'=U'+V'+T'\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\end{matrix}\)

\(\left(1\right)\Leftrightarrow U'=\dfrac{1}{2U}=\dfrac{1}{2\sqrt{x}}\)

(2) \(\Leftrightarrow V'=\dfrac{-1}{x^2.2V}=\dfrac{-1}{2x^2.\dfrac{1}{\sqrt{x}}}=\dfrac{-1}{2.\sqrt[3]{x^2}}\)

\(\left(3\right)\Leftrightarrow Y'=\dfrac{1}{2\sqrt{x}}-\dfrac{1}{2\sqrt[3]{x^2}}+x^9\)

ĐKXĐ:

a/ \(cos2x\ne0\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)

b/ \(sin\left(2x-7\pi\right)\ne0\Leftrightarrow sin2x\ne0\)

\(\Leftrightarrow2x\ne k\pi\Leftrightarrow x\ne\frac{k\pi}{2}\)

c/ \(sin\left(4x+5\pi\right).cos\left(2x-3\pi\right)\ne0\)

\(\Leftrightarrow sin4x.cos2x\ne0\)

\(\Leftrightarrow sin4x\ne0\) (vì \(sin4x=2sin2x.cos2x\) đã bao hàm luôn \(cos2x\) trong đó)

\(\Leftrightarrow4x\ne k\pi\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(f\left(x\right)=x^3-3x^2+\left(2m-2\right)x+m-3\\ f'\left(x\right)=3x^2-6x+2m-2\\ \Delta'_{f'}=-6m+15\)

Phương trình có 3 nghiệm phân biệt \(\Leftrightarrow\left\{{}\begin{matrix}\Delta'_{f'}=-6m+15>0\\y_{CĐ}y_{CT}< 0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m< \frac{5}{2}\\\left[\left(\frac{4}{3}m-\frac{10}{3}\right)x_{CĐ}+\frac{5}{3}m-\frac{11}{3}\right]\left[\left(\frac{4}{3}m-\frac{10}{3}\right)x_{CT}+\frac{5}{3}m-\frac{11}{3}\right]\end{matrix}\right.\\ \left\{{}\begin{matrix}m< \frac{5}{2}\\32m^3+3m^2-534m+823< 0\end{matrix}\right.\left(k\right)}\)

Theo định lí Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2+x_3=3>0\\x_1x_2+x_2x_3+x_3x_1=2m-2\\x_1x_2x_3=3-m>0\end{matrix}\right.\)

Từ \(x_1< -1< x_2< x_3\Rightarrow\left(x_1+1\right)\left(x_2+1\right)\left(x_2+1\right)< 0\)

Và từ \(\left(x_1+1\right)\left(x_2+1\right)\left(x_3+1\right)< 0\), do Viet ở trên nên \(x_1< -1< x_2< x_3\)

Vậy phương trình có 3 nghiệm thỏa mãn yêu cầu đề bài \(\Leftrightarrow\left(x_1+1\right)\left(x_2+1\right)\left(x_3+1\right)< 0\\ \Leftrightarrow x_1x_2x_3+x_1x_2+x_2x_3+x_3x_1+x_1+x_2+x_3+1< 0\\ \Leftrightarrow3-m+2m-2+3+1< 0\Leftrightarrow m< -5\)

Với m<-5 thì thỏa mãn điều kiện (k) ở trên. Vậy -10<m<-5

1.

\(4\left(1-cos^23x\right)+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}-4=0\)

\(\Leftrightarrow-4cos^23x+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=-\frac{1}{2}\\cos3x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{2\pi}{9}+\frac{k2\pi}{3}\\x=\pm\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

2.

\(\Leftrightarrow\frac{\sqrt{3}-1}{2\sqrt{2}}sinx-\frac{\sqrt{3}+1}{2\sqrt{2}}cosx=-\frac{\sqrt{3}-1}{2\sqrt{2}}\)

\(\Leftrightarrow sin\left(x-\frac{5\pi}{12}\right)=-cos\left(\frac{5\pi}{12}\right)\)

\(\Leftrightarrow sin\left(x-\frac{5\pi}{12}\right)=sin\left(-\frac{\pi}{12}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5\pi}{12}=-\frac{\pi}{12}+k2\pi\\x-\frac{5\pi}{12}=\frac{13\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)

\(3tan^2x+8tanx+8\sqrt{3}-9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k2\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k2\pi\end{matrix}\right.\)

4.

\(\Leftrightarrow sin\left(x-120^0\right)=-cos\left(2x\right)=sin\left(2x-90^0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-90^0=x-120^0+k360^0\\2x-90^0=300^0-x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow...\)

5.

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x=\frac{1}{2}-\frac{1}{2}cos6x\)

\(\Leftrightarrow cos6x=cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=2x+k2\pi\\6x=-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: ...

a/ \(cotx=\frac{2}{3}\Leftrightarrow x=arccot\left(\frac{2}{3}\right)+k\pi\)

b/ \(tanx=-\frac{4}{3}\Leftrightarrow x=arctan\left(-\frac{4}{3}\right)+k\pi\)

c/ \(sinx=\frac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

d/ \(cotx=-\frac{1}{2}\Leftrightarrow x=arccot\left(-\frac{1}{2}\right)+k\pi\)

e/ \(cotx=1\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

f/ \(sin\left(x+\frac{\pi}{2}\right)=1\Leftrightarrow x+\frac{\pi}{2}=\frac{\pi}{2}+k2\pi\Leftrightarrow x=k2\pi\)

g/ \(3x=k\pi\Leftrightarrow x=\frac{k\pi}{3}\)

d/

\(\Leftrightarrow2cos^3x+2sinx-6sin^2x.cosx=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2+2tanx.\frac{1}{cos^2x}-6tan^2x=0\)

\(\Leftrightarrow1+tanx\left(1+tan^2x\right)-3tan^2x=0\)

\(\Leftrightarrow tan^3x-3tan^2x+tanx+1=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x-2tanx-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=1-\sqrt{2}\\tanx=1+\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)

c/

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(4+2tan^3x-3tanx.\frac{1}{cos^2x}=0\)

\(\Leftrightarrow2tan^3x-3tanx\left(1+tan^2x\right)+4=0\)

\(\Leftrightarrow-tan^3x-3tanx+4=0\)

\(\Leftrightarrow\left(1-tanx\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x+tanx+4=0\left(vn\right)\end{matrix}\right.\)

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