b)Thấy sinx=0 không là nghiệm của pt

Chia cả hai vế của pt cho \(sin^3x\) ta được:

\(\dfrac{1}{sin^2x}-4+\dfrac{cosx}{sinx}.\dfrac{1}{sin^2x}=0\)

\(\Leftrightarrow1+cot^2x-4+cotx\left(1+cot^2x\right)=0\)

\(\Leftrightarrow cot^3x+cot^2x+cotx-3=0\)

\(\Leftrightarrow cotx=1\)\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

Vậy...

k) Pt \(\Leftrightarrow\left(sinx-cosx\right)\left(sin^2x+sinx.cosx+cos^2x\right)=sinx+cosx\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(1+sinx.cosx\right)=sinx+cosx\)

\(\Leftrightarrow sin^2x.cosx-sinx.cos^2x-2cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin^2x-sinx.cosx-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\-sin^2x-sinx.cosx-2cos^2x=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{2}+k\pi\) (\(k\in Z\))

Vậy...

j)\(\left(sinx-sin^2x\right)\left(sinx+2cosx\right)=\sqrt{3}\left(1+sinx\right)\left(1-sinx\right)^2\)

\(\Leftrightarrow sinx\left(1-sinx\right)\left(sinx+2cosx\right)=\sqrt{3}\left(1-sin^2x\right)\left(1-sinx\right)\)

\(\Leftrightarrow\left(1-sinx\right)\left[sin^2x+2sinx.cosx-\sqrt{3}\left(1-sin^2x\right)\right]=0\)

\(\Leftrightarrow\left(1-sinx\right)\left[sin^2x+2sinx.cosx-\sqrt{3}cos^2x\right]=0\)

\(\Leftrightarrow\left(1-sinx\right)\left[sinx-\left(-1-\sqrt{1+\sqrt{3}}\right)cosx\right]\left[sinx-\left(-1+\sqrt{1+\sqrt{3}}\right)cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\left(-1-\sqrt{1+\sqrt{3}}\right)cosx\\sinx=\left(-1+\sqrt{1+\sqrt{3}}\right)cosx\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\tan=-1-\sqrt{1+\sqrt{3}}\\tan=-1+\sqrt{1+\sqrt{3}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=arc.tan\left(-1-\sqrt{1+\sqrt{3}}\right)+k\pi\\x=arc.tan\left(-1+\sqrt{1+\sqrt{3}}\right)+k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

Vậy...

12.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cosx+sinx+\dfrac{1}{cosx}+\dfrac{1}{sinx}+tanx+cotx=-2\)

\(\Leftrightarrow cosx+sinx+\dfrac{cosx+sinx}{cosx.sinx}+\dfrac{sin^2x+cos^2x+2sinx.cosx}{sinx.cosx}=0\)

\(\Leftrightarrow cosx+sinx+\dfrac{cosx+sinx}{cosx.sinx}+\dfrac{\left(sinx+cosx\right)^2}{sinx.cosx}=0\)

\(\Leftrightarrow\left(cosx+sinx\right)\left(1+\dfrac{1}{cosx.sinx}+\dfrac{sinx+cosx}{sinx.cosx}\right)=0\)

\(\Leftrightarrow\dfrac{\left(cosx+sinx\right)\left(sinx+1\right)\left(cosx+1\right)}{sinx.cosx}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx+sinx=0\\sinx+1=0\\cosx+1=0\end{matrix}\right.\)

TH1: \(cosx+sinx=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\)

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

\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\left(tm\right)\)

TH2: \(sinx+1=0\)

\(\Leftrightarrow sinx=-1\)

\(\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\left(l\right)\)

TH3: \(cosx+1=0\)

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\left(l\right)\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{4}+k\pi\)

14.

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

\(\Leftrightarrow sin^2\left(\dfrac{\pi}{6}-x\right)+4cos\left(\dfrac{\pi}{6}-x\right)-4=0\)

\(\Leftrightarrow cos^2\left(\dfrac{\pi}{6}-x\right)+4cos\left(\dfrac{\pi}{6}-x\right)-5=0\)

\(\Leftrightarrow\left[cos\left(\dfrac{\pi}{6}-x\right)-1\right].\left[cos\left(\dfrac{\pi}{6}-x\right)+5\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{6}-x\right)=1\\cos\left(\dfrac{\pi}{6}-x\right)=-5\left(l\right)\end{matrix}\right.\)

\(cos\left(\dfrac{\pi}{6}-x\right)=1\Leftrightarrow\dfrac{\pi}{6}-x=0\Leftrightarrow x=\dfrac{\pi}{6}\left(tm\right)\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{6}\)

11.

Đặt \(sinx+cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(\Rightarrow sin2x=2sinx.cosx=t^2-1\)

Phương trình đã cho tương đương:

\(3t+2t^2+1=0\)

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

TH1: \(t=-1\)

\(\Leftrightarrow sinx+cosx=-1\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=cos\dfrac{3\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

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

TH2: \(t=-\dfrac{1}{2}\)

\(\Leftrightarrow sinx+cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=arccos\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x-\dfrac{\pi}{4}=-arccos\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}\pm arccos\left(-\dfrac{1}{2\sqrt{2}}\right)+k2\pi\)

Vậy ...

13.

ĐK: \(x\ne\dfrac{\pi}{2}+\dfrac{k\pi}{2}\)

Đặt \(sinx+cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình đã cho tương đương:

\(\dfrac{sinx+cosx}{sinx.cosx}=2\sqrt{2}\)

\(\Leftrightarrow\dfrac{2t}{t^2-1}=2\sqrt{2}\)

\(\Leftrightarrow t=\sqrt{2}t^2-\sqrt{2}\)

\(\Leftrightarrow t=\sqrt{2}t^2-\sqrt{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{2}}{2}\\t=-\sqrt{2}\end{matrix}\right.\)

TH1: \(t=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sinx+cosx=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

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

TH2: \(t=-\sqrt{2}\)

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

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=-\sqrt{2}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{4}+k\pi\left(tm\right)\\x=-\dfrac{3\pi}{4}+k\pi\left(tm\right)\end{matrix}\right.\)

Vậy ...

a.

\(sinx=2sin\dfrac{x}{2}cos\dfrac{x}{2}=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}}{sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}}=\dfrac{2tan\dfrac{x}{2}}{1+tan^2\dfrac{x}{2}}\)

Pt tương đương: \(\dfrac{2tan\dfrac{x}{2}}{1+tan^2\dfrac{x}{2}}+tan\dfrac{x}{2}=2\)

Đặt \(tan\dfrac{x}{2}=t\) ta được:

\(\dfrac{2t}{1+t^2}+t=2\Leftrightarrow2t+t\left(1+t^2\right)=2+2t^2\)

\(\Leftrightarrow t^3-2t^2+3t-2=0\)

\(\Leftrightarrow\left(t-1\right)\left(t^2-t+2\right)=0\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow tan\dfrac{x}{2}=1\)

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

7.

\(\Leftrightarrow2\left(2cos^2x\right)^2=3+5\left(2cos^22x-1\right)\)

\(\Leftrightarrow2\left(cos2x+1\right)^2=10cos^22x-2\)

\(\Leftrightarrow2cos^22x-cos2x-1=0\)

Đặt \(cos2x=t\Rightarrow2t^2-t-1=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

9.

Đặt \(sinx+cosx=t\) với \(\left|t\right|\le\sqrt{2}\)

\(\Rightarrow t^2=sin^2x+cos^2x+2sinx.cosx=1+sin2x\)

\(\Rightarrow sin2x=t^2-1\)

Phương trình trở thành:

\(t+t^2-1+3=0\)

\(\Leftrightarrow t^2+t+2=0\) (vô nghiệm)

5.

ĐK: \(x\ne\pi+k2\pi\)

Đặt \(t=tan\dfrac{x}{2}\Rightarrow sinx=\dfrac{2t}{1+t^2}\)

Phương trình trở thành:

\(\dfrac{2t}{1+t^2}+t=2\)

\(\Leftrightarrow2t+t+t^3=2+2t^2\)

\(\Leftrightarrow t^3-2t^2+3t-2=0\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow tan\dfrac{x}{2}=1\)

\(\Leftrightarrow sin\dfrac{x}{2}=cos\dfrac{x}{2}\)

\(\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

7.

\(\Leftrightarrow8\left(\dfrac{1+cos2x}{2}\right)^2=3+5\left(2cos^22x-1\right)\)

Đặt \(cos2x=t\) ta được:

\(8\left(\dfrac{t+1}{2}\right)^2=3+5\left(2t^2-1\right)\)

\(\Leftrightarrow2t^2-t-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

C.     có cấu hình điện tử vòng ngoài với 4 điện tử (cùng lúc tạo nên 4 liên kết cộng hoá trị với nguyên tử khác).

Đáp án C

Phương trình hóa học:

          CH₃COOH + C₂H₅OH D CH₃COOC₂H₅ + H₂O

Ban đầu:     1                    1

Phản ứng:  2/3        2/3                   2/3            2/3

Cân bằng:  1/3        1/3                   2/3            2/3

Hằng số cân bằng:

Vì H = 80% (tính theo axit) nên n_axit(_pu) = 0,8 mol

    CH₃COOH + C₂H₅OH D CH₃COOC₂H₅ +H₂O

Ban đầu:    1                   x

Phản ứng:0,8à    0,8 à             0,8 à          0,8

Cân bằng:0,2      (x-0,8)              0,8               0,8

Hằng số cân bằng:

Đáp án C

Phương trình hóa học:

          CH₃COOH + C₂H₅OH D CH₃COOC₂H₅ + H₂O

Ban đầu:     1                     1

Phản ứng:  2/3        2/3                   2/3            2/3

Cân bằng:  1/3        1/3                   2/3            2/3

Hằng số cân bằng:

Vì H = 80% (tính theo axit) nên n_axit(_pu) = 0,8 mol

    CH₃COOH + C₂H₅OH D CH₃COOC₂H₅ +H₂O

Ban đầu:    1           x

Phản ứng:0,8à    0,8 à             0,8 à          0,8

Cân bằng:0,2      (x-0,8)              0,8               0,8

Hằng số cân bằng: