1: =>(x-4)(x+1)=0

=>x=4 hoặc x=-1

2: =>x²-5x+6=0

=>(x-2)(x-3)=0

=>x=2 hoặc x=3

3: =>7x²+7x-x-1=0

=>(x+1)(7x-1)=0

=>x=-1 hoặc x=1/7

\(y=2-2.\left(2sinx.cosx\right)=2-2sin2x\)

Do \(-1\le sin2x\le1\Rightarrow0\le y\le4\)

\(y_{min}=0\) khi \(sin2x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

\(y_{max}=4\) khi \(sin2x=-1\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(y=2\left(1-cos2x\right)-cos2x=2-3cos2x\)

Do \(-1\le cos2x\le1\Rightarrow-1\le y\le5\)

\(y_{min}=-1\) khi \(cos2x=1\Leftrightarrow x=k\pi\)

\(y_{max}=5\) khi \(cos2x=-1\Rightarrow x=\dfrac{\pi}{2}+k\pi\)

ĐKXĐ: \(cos2x\ge0\Rightarrow0\le cos2x\le1\)

\(\Rightarrow3.0+1\le y\le3.\sqrt{1}+1\)

\(\Rightarrow1\le y\le4\)

\(y_{min}=1\) khi \(cos2x=0\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

\(y_{max}=4\) khi \(cos2x=1\Rightarrow x=k\pi\)

\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

\(y=4\left(1-sin^2x\right)+2sinx+2=-4sin^2x+2sinx+6\)

Đặt \(sinx=t\in\left[-1;1\right]\Rightarrow y=f\left(t\right)=-4t^2+2t+6\)

\(-\dfrac{b}{2a}=\dfrac{1}{4}\in\left[-1;1\right]\)

\(f\left(-1\right)=0\) ; \(f\left(\dfrac{1}{4}\right)=\dfrac{25}{4}\); \(f\left(1\right)=4\)

\(\Rightarrow y_{max}=\dfrac{25}{4}\) khi \(sinx=\dfrac{1}{4}\)

\(y_{min}=0\) khi \(sinx=-1\)

Ta có: \(y=4cos^2x+2sinx+2=4-4sin^2x+2sinx+2=-4sin^2x+2sinx+6=-\left(4sin^2x-2sinx+\dfrac{1}{16}-\dfrac{1}{16}-6\right)=-\left(2sin^2x-\dfrac{1}{4}\right)^2+\dfrac{97}{16}\)

Ta có: \(-\left(2sin^2x-\dfrac{1}{4}\right)^2\le0\Rightarrow y\le\dfrac{97}{16}\)

Vậy \(y_{max}=\dfrac{97}{16}\)

Q = (1 - 2x)(x - 3)

= x - 3 - 2x² + 6x

= - 2x² + 5x - 3

= \(-2\left(x^2-\frac{5}{2}x+3\right)=-2\left(x^2-2.\frac{5}{4}.x+\frac{25}{16}+\frac{23}{16}\right)=-2\left(x-\frac{5}{4}\right)^2-\frac{23}{8}\le-\frac{23}{8}\)

Dấu "=" xảy ra <=> x  - 5/4 = 0

=> x = 1,25

Vậy Max Q = -23/8 <=> x = 1,25

Q = ( 1 - 2x )( x - 3 )

= x - 3 - 2x² + 6x

= -2x² + 7x - 3

= -2( x² - 7/2x + 49/16 ) + 25/8

= -2( x - 7/4 )² + 25/8 ≤ 25/8 ∀ x

Đẳng thức xảy ra <=> x - 7/4 = 0 => x = 7/4

=> MaxQ = 25/8 <=> x = 7/4

\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)