\(B=l7x-3l+l7x+3l\)

     = \(l3-7xl+l7x+3l\) \(\ge l3-7x+7x+3l=6\)

Vậy GTNN là 6 khi -7/3 <= x <= 7/3 

Ngu Người **** cho tui đi 

\(a,\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

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

\(=|2x-1|+|2x-3|\)

\(b,\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

\(=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)

\(=|7x-3|+|7x+3|\)

=.= hok tốt!!

\(A=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)

\(A=\left|7x-3\right|+\left|7x+3\right|=\left|3-7x\right|+\left|7x+3\right|\)

\(A\ge\left|3-7x+7x+3\right|=6\)

\(A_{min}=6\) khi \(\left(3-7x\right)\left(7x+3\right)\ge0\Rightarrow-\frac{3}{7}\le x\le\frac{3}{7}\)

a) P=\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

=\(\left|2x-1\right|+\left|2x-3\right|\)

=\(\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|=\left|2\right|=2\)

<=> \(P\ge2\)

Dấu "=" xảy ra <=> (2x-1)(3-2x)\(\ge0\)

<=> \(\frac{1}{2}\le x\le\frac{3}{2}\)

Vậy min P=2 <=>\(\frac{1}{2}\le x\le\frac{3}{2}\)

b)Tương tự ý a

ở câu a P=\(\sqrt{4x^2-4x+1}\)+\(\sqrt{4x^2-12x+9}\)nha các bn

a) Ta có: \(P=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

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

\(=\left|2x-1\right|+\left|2x-3\right|\)

\(=\left|2x-1\right|+\left| 3-2x\right|\ge\left|2x-1+3-2x\right|=\left|2\right|=2\)

Dấu '=' xảy ra khi \(\left(2x-1\right)\left(3-2x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)\left(3-2x\right)>0\\\left(2x-1\right)\left(3-2x\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1>0\\3-2x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1< 0\\3-2x< 0\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}2x-1=0\\3-2x=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\frac{1}{2}\\x< \frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \frac{1}{2}\\x>\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

Vậy: Giá trị nhỏ nhất của biểu thức \(P=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\) là 2 khi \(\frac{1}{2}\le x\le\frac{3}{2}\)

b) Ta có: \(Q=\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

\(=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)

\(=\left|7x-3\right|+\left|7x+3\right|\)

\(=\left|7x-3\right|+\left|-7x-3\right|\ge\left|7x-3-7x-3\right|=\left|-6\right|=6\)

Dấu '=' xảy ra khi \(\left(7x-3\right)\left(-7x-3\right)\ge0\)

\(\Leftrightarrow\frac{-3}{7}\le x< \frac{3}{7}\)

Vậy: ...

Câu 1: 

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)(1)

Trường hợp 1: x<1

(1) trở thành 1-x+2-x=3

=>3-2x=3

=>x=0(nhận)

Trường hợp 2: 1<=x<2

(1) trở thành x-1+2-x=3

=>1=3(loại)

Trường hợp 3: x>=2

(1) trở thành x-1+x-2=3

=>2x-3=3

=>2x=6

hay x=3(nhận)

\(49x^2+9-42x=0\)

\(\Leftrightarrow\left(7x-3\right)^2=0\)

\(\Leftrightarrow7x-3=0\)

\(\Leftrightarrow7x=3\)