help me, please

1. a . 3x² - 6x = 0

\(\Leftrightarrow3x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

b. x³ - 13x = 0

\(\Leftrightarrow x\left(x^2-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)

c. 5x ( x - 2001 ) - x + 2001 = 0

<=> 5x ( x - 2001 ) - ( x - 2001 ) = 0

\(\Leftrightarrow\left(5x-1\right)\left(x-2001\right)=0\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2001=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2001\end{cases}}\)

2. a. \(2x^2+4x-8=2\left(x+1\right)^2-10\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+1\right)^2-10\ge-10\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy GTNN của bt trên = - 10 <=> x = - 1

b. \(-x^2-8x+1=-\left(x+4\right)^2+17\)

Vì \(\left(x+4\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+4\right)^2+17\le17\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+4\right)^2=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

Vậy GTLN của bt trên = 17 <=> x = - 4

\(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\sqrt{\left(x-1\right)^2}+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|x-6\right|\right)\)

\(=\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\)

Ta có \(\hept{\begin{cases}\left|x-4\right|\ge0\forall x\\\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\end{cases}}\)

=> \(\left|x-4\right|+\left(\left|x-1\right|+\left|6-x\right|\right)\ge5\forall x\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-4=0\\\left(x-1\right)\left(6-x\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

=> MinA = 5 <=> x = 4

Ta có: \(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(\Rightarrow A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)

\(=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)

\(=\left|x-4\right|+\left|x-1\right|+\left|x-6\right|\)

Xét \(\left|x-1\right|+\left|x-6\right|\)ta có: 

\(\left|x-1\right|+\left|x-6\right|=\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=\left|5\right|=5\)(1)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(6-x\right)\ge0\)

TH1: Nếu \(\hept{\begin{cases}x-1< 0\\6-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\6< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>6\end{cases}}\)( vô lý )

TH2: Nếu \(\hept{\begin{cases}x-1\ge0\\6-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\6\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le6\end{cases}}\Leftrightarrow1\le x\le6\)

mà \(\left|x-4\right|\ge0\)(2)

Từ (1) và (2) \(\Rightarrow A\ge5\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-4=0\\1\le x\le6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\1\le x\le6\end{cases}}\Leftrightarrow x=4\)

Vậy \(minA=5\)\(\Leftrightarrow x=4\)

Áp dụng bất đẳng thức giá trị tuyệt đối |a| + |b| \(\ge\) |a + b| ta có:

A = |x - 2001| + |x - 1| = |x - 2001| + |1 - x| \(\ge\) |(x - 2001) + (1 - x)| = |-2000| = 2000

=> A nhỏ nhất là 2000 ; chẳng hạn tại x = 1