\(N=x+2y-\sqrt{2x-1}-5\sqrt{4y-3}+13\)

\(2N=2x+4y-2\sqrt{2x-1}-10\sqrt{4y-3}+26\)

\(=\left(2x-1-2\sqrt{2x-1}+1\right)+\left(4y-3-10\sqrt{4y-3}+25\right)+4\)

\(=\left(\sqrt{2x-1}-1\right)^2+\left(\sqrt{4y-3}-5\right)^2+4\ge4\)

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

Ta có: \(2x^2-4x+3\)

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

\(=2\left(x^2-2.x.1+1^2+\frac{1}{2}\right)\)

\(=2[\left(x-1\right)^2+\frac{1}{2}]\)

\(=2\left(x-1\right)^2+1\ge1\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}\ge\sqrt{1}\)

\(\Rightarrow\sqrt{2\left(x-1\right)^2+1}+3\ge3+\sqrt{1}=4\)

\(\Rightarrow MinA=4\Leftrightarrow x=1\)

\(A=\sqrt{x^2-6x+2y^2+4y+11}+\sqrt{x^2+2x+3y^2+6y+4}\)

\(=\sqrt{\left(x^2-6x+9\right)+2\left(y^2+2y+1\right)}+\sqrt{\left(x^2+2x+1\right)+3\left(y^2+2y+1\right)}\)

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

\(\ge\sqrt{\left(x-3\right)^2+0}+\sqrt{\left(x+1\right)^2+0}\)

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

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

\(=4\)

Dấu bằng xảy ra khi và chỉ khi : 

\(\left(y+1\right)^2=0\Leftrightarrow y+1=0\Leftrightarrow y=-1\)

\(\left(x-3\right)\left(x+1\right)\ge0\Leftrightarrow x^2-2x-3\ge0\Leftrightarrow\left(x-1\right)^2\ge4\Leftrightarrow\left|x-1\right|\ge2\Leftrightarrow x\ge3;x\le-1\)

Vậy GTNN của biểu thức là 4 khi  \(x\ge3\) hoặc \(x\le-1\) và \(y=-1\)

Bạn dùng min copski
 

mình có giải sai ở chỗ dấu bằng xảy khi x... nhé. Phải là như này :

\(\left(3-x\right)\left(x+1\right)\ge0\Leftrightarrow...\Leftrightarrow x=-1;x=3\)

1) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x=25\Leftrightarrow x=\dfrac{25}{2}\left(tm\right)\)

2) \(=\sqrt{\dfrac{\dfrac{1}{4}}{9}}=\dfrac{\dfrac{1}{2}}{3}=\dfrac{1}{6}\)

3) \(=\sqrt{225a^2}=15a\left(do.a\ge0\right)\)

4) \(=2y^2.\dfrac{x^2}{2\left|y\right|}=\left[{}\begin{matrix}x^2y\left(y>0\right)\\-x^2y\left(y< 0\right)\end{matrix}\right.\)