\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)

\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)

\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)

\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)

\("="\Leftrightarrow x=3\)

Ngại làm lần 2 quá bạn ơi 

Câu hỏi của Chuột yêu Gạo - Toán lớp 9 | Học trực tuyến

Ta có: \(\sqrt{x^2-2x+10}=\sqrt{x^2-2x+1+9}=\sqrt{\left(x-1\right)^2+9}\ge\sqrt{9}\ge3\)

          \(\sqrt{x^2+4x+5}=\sqrt{x^2+4x+4+1}=\sqrt{\left(x+2\right)^2+1}\ge\sqrt{1}\ge1\)

    \(\Rightarrow\)   \(\sqrt{x^2-2x+10}+\sqrt{x^2+4x+5}\ge1+3\ge4\)

Vậy GTNN của biểu thức là 4

Ta có:

\(P=\sqrt{4x^2-12x+9}+\sqrt{4x^2-8x+4}\)

\(=\sqrt{\left(2x\right)^2-2.2x.3+3^2}+\sqrt{\left(2x\right)^2-2.2x.2+2^2}\)

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

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

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

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-3\ge0\\2-2x\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)

Vậy Min_P = 1 \(\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)

\(P=\sqrt{4x^2-12x+9}+\sqrt{4x^2-8x+4}\)

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

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

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

\(P=\sqrt{4x^2-12x+9}+\sqrt{4x^2-8x+4}\)

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

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

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

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có :

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

Đẳng thức xảy ra khi \(ab\ge0\)

=> \(\left(3-2x\right)\left(2x-2\right)\ge0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}3-2x\ge0\\2x-2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}-2x\ge-3\\2x\ge2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)

2. \(\hept{\begin{cases}3-2x\le0\\2x-2\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}-2x\le-3\\2x\le2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( loại )

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

\(A=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)

\(A=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)

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

\(A=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)

\(A=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=\left|2\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow1\le x\le2\)

\(B=\sqrt{x^2+4x+4}+\sqrt{x^2+6x+9}\)

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

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

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

Dấu "=" xảy ra \(\Leftrightarrow-3\le x\le-2\)

\(MinA=0\Leftrightarrow7+4x-4x^2=0\Leftrightarrow x=\dfrac{1\pm2\sqrt{2}}{2}\)