\(A=-x^2-6x+1\)

\(A=-\left(x^2+6x-1\right)\)

\(A=-\left(x^2+6x+9-10\right)\)

\(A=-\left(x^2+2\cdot x\cdot3+3^2\right)+10\)

\(A=-\left(x+3\right)^2+10\)

Có: \(\left(x+3\right)^2\ge0\forall x\Rightarrow-\left(x+3\right)^2\le0\)

\(\Rightarrow-\left(x+3\right)^2+10\le10\)

\(\Rightarrow A\le10\)

Dấu "=" xảy ra khi \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: \(A_{min}=10\Leftrightarrow x=-3\)

Từ đề bài, ta có:

129 : a = y (dư 10)

<=> a>10

<=> (129 - 10) : a = y <=> 119 : a = y

<=> 119 ⋮ a

a ϵ Ư(119) 

a ϵ {1; 7; 17; 119}

mà a>10

=> a ϵ {17; 119}

Với a = 17

119 : a = 119 : 17 = 7 

<=> y = 7

Với a = 119

119 : a = 119 : 119 = 1

<=> y = 1

Vậy (a; y) ϵ [(17; 7); (119;1)}

3.(1/2 - x ) + 1/3 = 7/6 - x

3/2 - 3x + 1/3 = 7/6 - x

3/2 - 3x + 1/3 - 7/6 + x = 0

3/2 + 1/3 - 7/6 - 3x + x =0

2/3 - 2x = 0

2/3 = 2x

x = 2/3 : 2 = 1/3

c) \(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\)

\(\dfrac{2\left(x+1\right)}{3}-\dfrac{6}{3}\ge\dfrac{x-2}{2}\)

\(\dfrac{2x+2-6}{3}=\dfrac{2x-4}{3}\ge\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{2\left(2x-4\right)}{2\cdot3}\ge\dfrac{3\left(x-2\right)}{3\cdot2}\)

\(\Leftrightarrow\dfrac{4x-8}{6}\ge\dfrac{3x-6}{6}\)

\(\Leftrightarrow4x-8\ge3x-6\)

\(\Leftrightarrow4x-8-3x+6\ge3x-6-3x+6\)

\(\Leftrightarrow x-2\ge0\Leftrightarrow x\ge0+2=2\)

a)\(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\)

\(\dfrac{x\cdot2x}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=5\)

\(\dfrac{2x^2}{x^2-x}+\dfrac{3\left(x^2-1^2\right)}{x^2-x}=5\)

\(\dfrac{2x^2}{x^2-x}+\dfrac{3x^2-3}{x^2-x}=5\)

\(\dfrac{2x^2+3x^2-3}{x^2-x}=\dfrac{5x^2-3}{x^2-x}=5\)

\(\Rightarrow5x^2-3=5\left(x^2-x\right)=5x^2-5x\)

\(\Rightarrow3=5x\)

\(x=\dfrac{3}{5}\)

b) \(\left|1-2x\right|=2x-1\)

TH1: \(1>2x\)

\(\Rightarrow\left[{}\begin{matrix}1-2x>0\\2x-1< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left|1-2x\right|>0\\2x-1< 0\end{matrix}\right.\) => Vô lí

TH2: \(1\le2x\)

\(\Rightarrow\left[{}\begin{matrix}1-2x\le0\Rightarrow\left|1-2x\right|\ge0\\2x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|1-2x\right|=2x-1\ge0\)

\(\Leftrightarrow2x-1\ge0\Rightarrow2x-1+1=2x\ge0+1=1\)

\(\Leftrightarrow\dfrac{2x}{2}=x\ge\dfrac{1}{2}\)

\(B=\left|x-3\right|+\left|x-5\right|\)

Vì \(\left|x-3\right|,\left|x-5\right|\ge0\forall x\)

\(\Rightarrow\left|x-3\right|+\left|x-5\right|\ge0\forall x\)

\(\Rightarrow B\ge0\)

+) Nếu \(x< 3\Rightarrow\left|x-3\right|=3-x,\left|x-5\right|=5-x\) 

\(\Rightarrow B=\left|x-3\right|+\left|x-5\right|=3-x+5-x=8-2x\)

TH1: Nếu \(x< 0\Rightarrow-x>0\Rightarrow B=8+2\left(-x\right)>8\)

TH2: Nếu \(x=0\Rightarrow B=8\)

TH3: Nếu \(0< x< 3\Rightarrow B=8-2x< 8\)

+) Nếu \(x=3\Rightarrow\left|x-3\right|=0,\left|x-5\right|=2\) 

\(\Rightarrow B=\left|x-3\right|+\left|x-5\right|=2\)

+) Nếu \(3< x< 5\Rightarrow\left|x-3\right|=x-3,\left|x-5\right|=5-x\) 

\(\Rightarrow B=\left|x-3\right|+\left|x-5\right|=x-3+5-x=2\)

+) Nếu \(x=5\Rightarrow\left|x-3\right|=2,\left|x-5\right|=0\)

\(\Rightarrow B=\left|x-3\right|+\left|x-5\right|=2\)

+) Nếu \(x>5\Rightarrow\left|x-3\right|=x-3,\left|x-5\right|=x-5\)

\(B=\left|x-3\right|+\left|x-5\right|=x-3+x-5=2x-8>2\)

\(\Rightarrow B_{max}>8\Leftrightarrow x< 0\)

\(-4x^3+4x=0\)

Áp dụng công thức phương trình bậc 3, ta có:

\(a=-4,b=0,c=4,d=0\)

\(\Rightarrow\Delta=b^2-3ac=0^2-3\cdot-4\cdot4=0+48=48\)

\(\Rightarrow k=\dfrac{9abc-2b^3-27a^2d}{2\sqrt{\left|\Delta\right|^3}}\)

\(\Rightarrow k=\dfrac{9\cdot-4\cdot0\cdot4-2\cdot0^3-27\cdot\left(-4\right)^2\cdot0}{2\sqrt{\left|48\right|^3}}\)

\(\Rightarrow k=\dfrac{0}{2\sqrt{\left|48\right|^3}}=0\)

Vì Δ = 48 > 0 và k = 0 < 1

\(\Rightarrow x_1=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}\right)-b}{3a}\)

\(x_1=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)}{3}\right)-0}{3\cdot-4}\)

\(x_1=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}}{3}\right)}{-12}\)

\(x_1=\dfrac{8\sqrt{3}cos\left(\dfrac{\pi}{6}\right)}{-12}\)

\(x_1=\dfrac{8\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}}{-12}\)

\(x_1=\dfrac{\dfrac{8\sqrt{3}\cdot\sqrt{3}}{2}}{-12}\)

\(x_1=\dfrac{4\cdot3}{-12}=\dfrac{12}{-12}=-1\)

\(\Rightarrow x_2=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}-\dfrac{2\pi}{3}\right)-b}{3a}\)

\(x_2=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)-2\pi}{3}\right)-0}{3\cdot-4}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{arccos\left(0\right)-2\pi}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}-2\pi}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{-3\pi}{2}}{3}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}cos\left(\dfrac{-3\pi}{6}\right)}{-12}=\dfrac{8\sqrt{3}cos\left(\dfrac{-\pi}{2}\right)}{-12}\)

\(x_2=\dfrac{8\sqrt{3}\cdot0}{-12}=0\)

\(\Rightarrow x_3=\dfrac{2\sqrt{\Delta}cos\left(\dfrac{arccos\left(k\right)}{3}+\dfrac{2\pi}{3}\right)-b}{3a}\)

\(x_3=\dfrac{2\sqrt{48}cos\left(\dfrac{arccos\left(0\right)+2\pi}{3}\right)-0}{3\cdot-4}\)

\(x_3=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{\pi}{2}+2\pi}{3}\right)}{-12}=\dfrac{8\sqrt{3}cos\left(\dfrac{\dfrac{5\pi}{2}}{3}\right)}{-12}\)

\(x_3=\dfrac{8\sqrt{3}cos\left(\dfrac{5\pi}{6}\right)}{-12}=\dfrac{8\sqrt{3}\cdot\dfrac{-\sqrt{3}}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{8\sqrt{3}\cdot-\sqrt{3}}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{8\cdot-3}{2}}{-12}\)

\(x_3=\dfrac{\dfrac{-24}{2}}{-12}\)

\(x_3=\dfrac{-12}{-12}=1\)

Vậy: \(x_1=-1,x_2=0,x_3=1\)