1A

2C

3C

Áp dụng hệ thức lượng trong tam giác vuông có:

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{61}{900}\)\(\Leftrightarrow AH^2=\dfrac{900}{61}\)\(\Leftrightarrow AH=\dfrac{30\sqrt{61}}{61}\)

Áp dụng BĐT:

\(\sqrt{x^2+a^2+u^2}+\sqrt{y^2+b^2+w^2}\ge\sqrt{\left(x+y\right)^2+\left(a+b\right)^2+\left(u+w\right)^2}\) (1)

Thật vậy (1)\(\Leftrightarrow x^2+a^2+u^2+y^2+b^2+w^2+2\sqrt{\left(x^2+a^2+u^2\right)\left(y^2+b^2+w^2\right)}\ge x^2+2xy+y^2+a^2+2ab+b^2+u^2+2uw+w^2\)

\(\Leftrightarrow\sqrt{x^2y^2+x^2b^2+x^2w^2+a^2y^2+a^2b^2+a^2w^2+u^2y^2+u^2b^2+u^2w^2}\ge xy+ab+uw\)

\(\Leftrightarrow x^2y^2+x^2b^2+x^2w^2+a^2y^2+a^2b^2+a^2w^2+u^2y^2+u^2b^2+u^2w^2\ge x^2y^2+a^2b^2+u^2w^2+2xyab+2abuw+2xyuw\)

\(\Leftrightarrow\left(x^2b^2-2xyab+a^2y^2\right)+\left(x^2w^2-2xyuw+u^2y^2\right)+\left(a^2w^2-2abuw+u^2b^2\right)\ge0\)

\(\Leftrightarrow\left(xb-ay\right)^2+\left(xw-uy\right)^2+\left(aw-ub\right)^2\ge0\) (lđ)

Dấu "=" xảy ra khi \(\dfrac{x}{y}=\dfrac{a}{b}=\dfrac{u}{w}\)

Áp dụng (1) vào A ta được:

\(A\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2}+\sqrt{c^2+\dfrac{1}{c^2}+\dfrac{1}{a^2}}\)\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}\right)^2}\)

\(\Leftrightarrow A\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}\)

\(\Leftrightarrow A\ge\dfrac{3\sqrt{33}}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)

Vậy...

Câu 6:C

Câu 8:C

Câu 9:Tìm phần bù của B trong A có nghĩa là tìm A\B

Ý D

à nhầm, \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{3}\\x=\dfrac{\pi}{3}\end{matrix}\right.\)

11. 

\(\left(2cosx+5\right)\left(sin^4\dfrac{x}{2}-cos^4\dfrac{x}{2}\right)+3=0\)

\(\Leftrightarrow\left(2cosx+5\right)\left(sin^2\dfrac{x}{2}-cos^2\dfrac{x}{2}\right)+3=0\)

\(\Leftrightarrow\left(2cosx+5\right)\left(-cosx\right)+3=0\)

\(\Leftrightarrow-2cos^2x-5cosx+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cosx=-3\left(vn\right)\end{matrix}\right.\)\(\Rightarrow x=\pm\dfrac{\pi}{3}+k2\pi\) \(\left(k\in Z\right)\)

\(x\in\left(0;2\pi\right)\Rightarrow\)\(\left[{}\begin{matrix}0< -\dfrac{\pi}{3}+k2\pi< 2\pi\\0< \dfrac{\pi}{3}+k2\pi< 2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{6}< k< \dfrac{7}{6}\\-\dfrac{1}{6}< k< \dfrac{5}{6}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}\\x=\dfrac{\pi}{3}\end{matrix}\right.\)

Vậy...

14)\(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)

\(\Leftrightarrow\sqrt{3}sin3x+cos3x=sinx+\sqrt{3}cosx\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin3x+\dfrac{1}{2}cos3x=\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\)

\(\Leftrightarrow sin3x.cos\dfrac{\pi}{6}+cos3x.sin\dfrac{\pi}{6}=sinx.cos\dfrac{\pi}{3}+cosx.\dfrac{\pi}{3}\)

\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{6}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)

\(b+d=\dfrac{\pi}{6}+\dfrac{\pi}{3}=\dfrac{\pi}{2}\)

Vậy...

CT: \(asinx+bcosx=c\) (đối với pt có nghiệm)

\(\Leftrightarrow\dfrac{a}{\sqrt{a^2+b^2}}sinx+\dfrac{b}{\sqrt{a^2+b^2}}cosx=\dfrac{c}{\sqrt{a^2+b^2}}\)

Do \(\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\dfrac{b}{\sqrt{a^2+b^2}}\right)^2=1\)

\(\Rightarrow\exists\alpha:\left\{{}\begin{matrix}cos\alpha=\dfrac{a}{\sqrt{a^2+b^2}}\\sin\alpha=\dfrac{b}{\sqrt{a^2+b^2}}\end{matrix}\right.\)

\(\Rightarrow c=\sqrt{a^2+b^2}\left(sinx.cos\alpha+cosx.sin\alpha\right)\)

\(=\sqrt{a^2+b^2}sin\left(x+\alpha\right)\)

(Bạn áp dụng công thức này vào làm,nhiều quá nên lười)

11. \(cos^2x-sin2x=\sqrt{2}+cos^2\left(\dfrac{\pi}{2}+x\right)\)

\(\Leftrightarrow cos^2x-sin2x=\sqrt{2}+sin^2x\)

\(\Leftrightarrow cos^2x-sin^2x-sin2x=\sqrt{2}\)

\(\Leftrightarrow-sin2x+cos2x=\sqrt{2}\)

\(\Leftrightarrow-\dfrac{1}{\sqrt{2}}.sin2x+\dfrac{1}{\sqrt{2}}cos2x=1\)

\(\Leftrightarrow cos2x.cos\dfrac{\pi}{4}-sin2x.sin\dfrac{\pi}{4}=1\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow2x+\dfrac{\pi}{4}=k2\pi\left(k\in Z\right)\)

\(\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)

Do \(x\in\left(0;3\pi\right)\)\(\Rightarrow0< -\dfrac{\pi}{8}+k\pi< 3\pi\left(k\in Z\right)\)

\(\Rightarrow\dfrac{1}{8}< k< \dfrac{25}{8}\left(k\in Z\right)\)\(\Rightarrow k\in\left\{1;2;3\right\}\)

\(\Rightarrow x\in\left\{\dfrac{7\pi}{8};\dfrac{15\pi}{8};\dfrac{23\pi}{8}\right\}\)

Vậy...

b dương :) ,cười cái mả cha ấy.

Mai trần Vì \(b\le-b\) là vô lí