a, \(y=2sin^2x-cos2x=1-2cos2x\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=2sin^2x-cos2x\in\left[-1;3\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=-1\\y_{max}=3\end{matrix}\right.\)

Giúp mình với 

Áp dụng BĐT cosi:

\(A=\left(3x+\dfrac{3}{x}\right)+\left(\dfrac{4}{9}y+\dfrac{4}{y}\right)+\left(2x+y\right)\\ A\ge2\sqrt{\dfrac{9x}{x}}+2\sqrt{\dfrac{16y}{9y}}+5\\ A\ge2\cdot3+2\cdot\dfrac{4}{3}+5=\dfrac{41}{3}\)

Vậy \(A_{min}=\dfrac{41}{3}\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{3}{x}\\\dfrac{4y}{9}=\dfrac{4}{y}\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)

\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)

Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)

\(y=\sqrt{5}sin\left(2x-a\right)+7\)

\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)

b: 

3/2pi<x<2pi

=>cosx>0; sin x<0

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

=>\(\dfrac{1}{cos^2x}=1+\left(-3\right)^2=10\)

=>cosx=1/căn 10

=>sin x=-3/căn 10

\(A=\sqrt{10}\cdot\dfrac{1}{\sqrt{10}}-2\cdot\dfrac{-3}{\sqrt{10}}+3=4+\dfrac{6}{\sqrt{10}}=\dfrac{4\sqrt{10}+6}{\sqrt{10}}\)

a: cot x=3 nên cosx/sinx=3

=>cosx=3*sinx

\(B=\dfrac{2sin^2x+3sinx\cdot3\cdot sinx}{1-2\cdot\left(3\cdot sinx\right)^2}=\dfrac{11sin^2x}{sin^2x+cos^2x-18sin^2x}\)

\(=\dfrac{11sin^2x}{-17sin^2x+9sin^2x}=\dfrac{-11}{8}\)

a.

\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)

Dấu "=" xảy ra khi \(x=2013\)

b.

\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)

\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)

\(B=\dfrac{-2x^2-1+2x^2+4x+2}{4x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+1\right)^2}{2x^2+1}\ge-\dfrac{1}{2}\)

\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)

Đk: \(x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

PT \(\Leftrightarrow2\left(sinx.\dfrac{\sqrt{2}}{2}-cosx.\dfrac{\sqrt{2}}{2}\right)^2=2sin^2x-\dfrac{sinx}{cosx}\)

\(\Leftrightarrow\left(sinx-cosx\right)^2=2sin^2x-\dfrac{sinx}{cosx}\)

\(\Leftrightarrow1-2.sinx.cosx=2sin^2x-\dfrac{sinx}{cosx}\)

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

\(\Leftrightarrow\left(cosx+sinx\right)-2sinx.cosx\left(cosx+sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx+sinx=0\\1-2sinx.cosx=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\sin2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) ( k nguyên ) (tmđk)

Vậy...

𝐼 𝒹𝑜𝓃'𝓉 𝒸𝒶𝓇𝑒 𝒽𝑜𝓌 𝓁𝑜𝓃𝑔 𝒾𝓉 𝓉𝒶𝓀𝑒𝓈

Ta có: \(\left(x-1\right)^2+\left(x+y\right)^2\le9\Rightarrow x+y\le3\).

Áp dụng bất đẳng thức AM - GM ta có:

\(\dfrac{2}{x}+2x\ge2\sqrt{\dfrac{2}{x}.2x}=4;\dfrac{4}{y}+y\ge2\sqrt{\dfrac{4}{y}.y}=4\).

Do đó \(\dfrac{2}{x}\ge4-2x;\dfrac{4}{y}\ge4-y\)

\(\Rightarrow P\ge8-4\left(x+y\right)\ge-4\). (do \(x+y\le3\)).

Vậy...

Đẳng thức xảy ra khi và chỉ khi x = 1; y = 2.