\(\dfrac{1}{cos^2\alpha}=1+tan^2\alpha=1+\left(\dfrac{7}{24}\right)^2=\dfrac{625}{576}\)

\(\Rightarrow cos^2\alpha=\dfrac{576}{625}\)

\(S_{xq}=\pi.r.l\Leftrightarrow235,5=\pi.10.l\Leftrightarrow l=\dfrac{235,5}{10\pi}\approx7,496\left(cm\right)\)

\(xy+yz+zx=3xyz\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)

Có \(\dfrac{1}{x+2y+3z}=\dfrac{1}{\left(x+y\right)+\left(y+z\right)+2z}\le\dfrac{1}{9}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{2z}\right)\le\dfrac{1}{9}\left(\dfrac{1}{4x}+\dfrac{1}{4y}+\dfrac{1}{4y}+\dfrac{1}{4z}+\dfrac{1}{2z}\right)=\dfrac{1}{9}\left(\dfrac{1}{4x}+\dfrac{1}{2y}+\dfrac{3}{4z}\right)\)

Tương tự cx có: \(\dfrac{1}{y+2z+3x}\le\dfrac{1}{9}\left(\dfrac{1}{4y}+\dfrac{1}{2z}+\dfrac{3}{4x}\right)\);\(\dfrac{1}{z+2x+3y}\le\dfrac{1}{9}\left(\dfrac{1}{4z}+\dfrac{1}{2x}+\dfrac{3}{4y}\right)\)

Cộng vế với vế \(\Rightarrow\Sigma\dfrac{1}{x+2y+3z}\le\dfrac{1}{9}\left(\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{3}{4}\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{2}\)

Dấu "=" xayra khi x=y=z=1

Vậy \(P_{max}=\dfrac{1}{2}\)

\(\left|x+2\right|+\left|x-2\right|=3\)

TH1:\(x< -2\)

Pt \(\Leftrightarrow-\left(x+2\right)-\left(x-2\right)=3\)

\(\Leftrightarrow x=-\dfrac{3}{2}\left(ktm\right)\)

TH2:\(-2\le x\le2\)

Pt \(\Leftrightarrow x+2-\left(x-2\right)=3\Leftrightarrow4=3\) (vô lí)

TH3:\(>2\)

Pt \(\Leftrightarrow x+2+x-2=3\Leftrightarrow x=\dfrac{3}{2}\left(ktm\right)\)

Vậy pt vô nghiệm

\(\dfrac{2}{cos^2x}=\dfrac{2\left(cos^2x+sin^2x\right)}{cos^2x}=2+\dfrac{2sin^2x}{cos^2x}=2+2tan^2x=2\left(1+tan^2x\right)\)

1.Ý C

Hàm số có nghĩa khi \(x^2+14x+45\ne0\Leftrightarrow x\ne\left\{-5;-9\right\}\)

\(\Rightarrow D=R\backslash\left\{-5;-9\right\}\)

2. Ý D

Hàm số có nghĩa khi \(\left\{{}\begin{matrix}x+7\ge0\\x^2+6x-16\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-7\\x\ne\left\{2;-8\right\}\end{matrix}\right.\)

\(\Rightarrow D=\)\([-7;+ \infty) \)\(\backslash\left\{2\right\}\)

\(y=\left|sinx-\left(1-2sin^2x\right)\right|=\left|2sin^2x+sinx-1\right|\)

Đặt \(t=sinx;-1\le t\le1\)

\(\Rightarrow y=\left|2t^2+t-1\right|\)

Đặt \(f\left(t\right)=2t^2+t-1;-1\le t\le1\)

Vẽ BBT của \(f\left(t\right)=2t^2+t-1;-1\le t\le1\) sẽ tìm được \(f\left(t\right)_{min}=-\dfrac{9}{8};f\left(t\right)_{max}=2\)

\(\Rightarrow0\le\left|f\left(t\right)\right|\le2\)

\(\Leftrightarrow0\le y\le2\)

\(\Rightarrow y_{min}=0\Leftrightarrow2sin^2x+sinx-1=0\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(y_{max}=2\Leftrightarrow t=1\Leftrightarrow sinx=1\)

1)Do \(\pi< \alpha< \dfrac{3\pi}{2}\)\(\Rightarrow sin\alpha< 0\)

\(sin\alpha=-\sqrt{1-cos^2\alpha}=-\dfrac{\sqrt{7}}{4}\)

\(cos2\alpha=2cos^2\alpha-1=\dfrac{1}{8}\)

2)\(\dfrac{1-cosx+cos2x}{sin2x-sinx}=\dfrac{2cos^2x-cosx}{2sinx.cosx-sinx}=\dfrac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\dfrac{cosx}{sinx}=cotx\)

\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{24}{7}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{576}{625}\Rightarrow cos\alpha=\dfrac{24}{25}\)

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow sin^2\alpha=\dfrac{49}{625}\Rightarrow cos\alpha=\dfrac{7}{25}\)

Áp dụng hẹ thức lượng trong tam giác vuông:

\(AB.AC=AH.BC=78\)

\(\Rightarrow AB=\dfrac{78}{AC}\)

Lại có:\(AB^2+AC^2=BC^2\Leftrightarrow\left(\dfrac{78}{AC}\right)^2+AC^2=169\)

\(\Leftrightarrow AC^4-169AC^2+6084=0\)\(\Leftrightarrow\left[{}\begin{matrix}AC=\sqrt{117}=3\sqrt{13}\\AC=\sqrt{52}=2\sqrt{13}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}AB=2\sqrt{13}\\AB=3\sqrt{13}\end{matrix}\right.\)

Vậy \(AB=2\sqrt{13};AC=3\sqrt{13}\) hoặc \(AC=2\sqrt{13};AB=3\sqrt{13}\)