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

Đề bài : Tìm Min của \(D=\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

Ta có ; \(D=\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}=\sqrt{49\left(x-\frac{3}{7}\right)^2}+\sqrt{49\left(x+\frac{3}{7}\right)^2}=7\left(\left|x-\frac{3}{7}\right|+\left|x+\frac{3}{7}\right|\right)\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\). Dấu "=" xảy ra khi a,b cùng dấu.

Được; \(D=7\left(\left|\frac{3}{7}-x\right|+\left|x+\frac{3}{7}\right|\right)\ge7.\left|\frac{3}{7}-x+x+\frac{3}{7}\right|=7.\frac{6}{7}=6\)

\(\Rightarrow D\ge6\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+\frac{3}{7}\ge0\\\frac{3}{7}-x\ge0\end{cases}\Leftrightarrow}\frac{-3}{7}\le x\le\frac{3}{7}\)

Vậy Min D = 6 \(\Leftrightarrow\frac{-3}{7}\le x\le\frac{3}{7}\)

Mình thấy đề bài hơi kì kì ^^

Ta có ; \(D=2\sqrt{49x^2-42x+9}=2\sqrt{49\left(x-\frac{3}{7}\right)^2}=14\left|x-\frac{3}{7}\right|\ge0\)

Do đó Min D = 0 \(\Leftrightarrow x=\frac{3}{7}\)

viết lại đề bài: \(\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

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: ...

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

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

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

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

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

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

=.= 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 \(\left(3-7x\right)\left(7x+3\right)\ge0\Rightarrow-\frac{3}{7}\le x\le\frac{3}{7}\)

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)

\(49x^2-22x+9=\left(7x\right)^2-2.7.\dfrac{11}{7}x+\dfrac{121}{49}+\dfrac{320}{49}\)

\(=\left(7x-\dfrac{11}{7}\right)^2+\dfrac{320}{49}\ge\dfrac{320}{49}\) dấu"=" xảy ra<=>\(x=\dfrac{11}{49}\)

\(=>\sqrt{49x^2-22x+9}\ge\)\(\sqrt{\dfrac{320}{49}}=\dfrac{8\sqrt{5}}{7}\)

\(=>B\ge\dfrac{8\sqrt{5}}{7}+8\sqrt{38}\)