28,Pt\(\Leftrightarrow4\left(1-2sin^22x\right)+10sin2x-7=0\)

\(\Leftrightarrow-8sin^22x+10sin2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\dfrac{3}{4}\\sin2x=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}.arc.sin\left(\dfrac{3}{4}\right)+k\pi\\x=\dfrac{\pi}{2}-arc.sin\left(\dfrac{3}{4}\right)+k\pi\\x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy..

30,\(sin\left(x+\dfrac{\pi}{4}\right)=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)

\(x\in\left[\pi;5\pi\right]\)\(\Rightarrow\pi\le\dfrac{\pi}{2}+k2\pi\le5\pi\)\(\left(k\in Z\right)\)\(\Leftrightarrow\dfrac{1}{4}\le k\le\dfrac{9}{4}\)\(\left(k\in Z\right)\)

\(\Rightarrow k=\left\{1;2\right\}\)

\(\Rightarrow x=\dfrac{5\pi}{2};x=\dfrac{9\pi}{2}\)

Có \(\dfrac{sin\alpha}{cos\alpha}=tan\alpha=2\)\(\Rightarrow sin\alpha=2cos\alpha\)

\(\dfrac{sin\alpha+cos\alpha}{sin\alpha-cos\alpha}=\dfrac{2cos\alpha+cos\alpha}{2cos\alpha-cos\alpha}=\dfrac{3cos\alpha}{cos\alpha}=3\)

\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Áp dụng bđt Cauchy-Schwarz dạng Engel có:

\(A\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{2}}\ge\dfrac{4}{1}+\dfrac{1}{\dfrac{1}{2}}=6\)

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

Vậy GTNN của A=6

\(S=\dfrac{1}{a^3+b^3}+\dfrac{\dfrac{9}{4}}{3a^2b}+\dfrac{\dfrac{9}{4}}{3ab^2}+\dfrac{1}{4ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Áp dụng bđt Cauchy-Schwarz dạng Engel có:

\(S\ge\dfrac{\left(1+\dfrac{3}{2}+\dfrac{3}{2}\right)^2}{a^3+3a^2b+3ab^2+b^3}+\dfrac{1}{4ab}.\dfrac{4}{a+b}\)

\(\Leftrightarrow S\ge\dfrac{16}{\left(a+b\right)^3}+\dfrac{1}{\left(a+b\right)^2}.\dfrac{4}{a+b}\)

\(\Leftrightarrow S\ge\dfrac{16}{1}+\dfrac{1}{1}.\dfrac{4}{1}=20\)

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

Vậy GTNN của \(S=20\) khi \(a=b=\dfrac{1}{2}\)

Ý A

\(f\left(0\right)+2f\left(7\right)-g\left(1\right)=0^2-1+2\sqrt{7+2}-\left(1^2+2\right)=2\)

\(E\cup F=R\)

\(\Leftrightarrow4-a< 2a\Leftrightarrow a>\dfrac{4}{3}\)

Vậy \(a>\dfrac{4}{3}\)

Đk:\(x^2-4\ge0\)

Pttt:\(\Leftrightarrow\sqrt{\left(x^2-4\right)+4\sqrt{x^2-4}+4}=x^2-4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x^2-4}+2\right)^2}=x^2-4\)

\(\Leftrightarrow\sqrt{x^2-4}+2=x^2-4\)

\(\Leftrightarrow\left(x^2-4\right)-\sqrt{x^2-4}-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4}=2\\\sqrt{x^2-4}=-1\left(vn\right)\end{matrix}\right.\)\(\Rightarrow x^2-4=4\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\) (tm)

Vậy...

Thiếu đề ko e?

Xét \(3a^4-4a^3+1=\left(a-1\right)^2\left(3a^2+2a+1\right)\ge0\)

\(2b^3-3b^2+1=\left(2b+1\right)\left(b-1\right)^2\ge0;\forall b>0\)

\(c^2-2c+1=\left(c-1\right)^2\ge0\)

\(\Rightarrow\left(3a^4+2b^3+c^2\right)-\left(4a^3+3b^2+2c\right)+3\ge0\)

\(\Rightarrow3a^4+2b^3+c^2\ge4a^3+3b^2+2c-3=6\)

Dấu "=" xảy ra khi a=b=c=1

ừ,chị nhầm :D