Trước hết bằng phép biến đổi tương đương ; ta chứng minh bất đẳng thức phụ sau:

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}...\)

Biểu diễn: 

\(y=\sqrt{2}\left(\sqrt{x^2-x+\frac{5}{2}}+\sqrt{x^2-2x+2}\right)\)

  \(=\sqrt{2}\left(\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{9}{4}}+\sqrt{\left(1-x\right)^2+1}\right)\)

  \(\ge\sqrt{2}\sqrt{\left(x-\frac{1}{2}+1-x\right)^2+\left(\frac{3}{2}+1\right)^2}=\sqrt{13}.\)

Vậy giá trị nhỏ nhất của \(y=\sqrt{13}\Leftrightarrow x=\frac{4}{5}.\)

Trong căn thứ nhất là 4x ms đúng chứ nhỉ

Nếu đề bài là 4x thì cách giải nè :

2x² + 4x + 3 = 2.(x² + 2x +1) + 1 = 2.(x+1)² + 1 >= 1 ( >= là dấu lớn hơn hoặc bằng )  khi đó căn thứ nhất >= căn 1 =1

x² + 2x + 3 = (x+1)² + 2 >=2  khi đó căn thứ 2 >= căn 2

Suy ra y>= 1 + căn 2

Dấu = xảy ra khi x+1=0 khi x=-1

À mình bấm nhầm 4x mới đúng. Cảm ơn bạn

a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)

\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2

\(B=\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}=\sqrt{2\left[x^2-2x+1+\dfrac{3}{2}\right]}\)

\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1

\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1

\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2

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

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

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)

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

Vì \(\left(2x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+1\right)^2+1\ge1\forall x\)

\(\Rightarrow A\ge\sqrt{1}=1\)

Dấu " = " xảy ra \(\Leftrightarrow2x+1=0\)\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(minA=1\Leftrightarrow x=\frac{-1}{2}\)

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

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

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-1\right)^2+4\ge4\forall x\)

\(\Rightarrow B\ge\sqrt{4}=2\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(minB=2\Leftrightarrow x=1\)

Mơn bạn nha

1.

Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$A=|x+2|+|x+3|=|x+2|+|-x-3|\geq |x+2-x-3|=1$

Vậy GTNN của $A$ là $1$. Giá trị này đạt tại $(x+2)(-x-3)\geq 0$

$\Leftrightarrow (x+2)(x+3)\leq 0$

$\Leftrightarrow -3\leq x\leq -2$

2. ĐKXĐ: $x\geq 1$

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

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

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

Vậy gtnn của $B$ là $2$. Giá trị này đạt tại $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$

$\Leftrightarrow 1-\sqrt{x-1}\geq 0$

$\Leftrightarrow 0\leq x\leq 2$

3.

$C\sqrt{2}=\sqrt{4x+2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}$

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

Vì $\sqrt{4x-1}\geq 0$ nên $|\sqrt{4x-1}+1|\geq 1$

$\Rightarrow C\sqrt{2}\geq 2$

$\Rightarrow C\geq \sqrt{2}$

Vậy $C_{\min}=\sqrt{2}$. Giá trị này đạt tại $x=\frac{1}{4}$

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

Ta có: \(\sqrt{2x^2-4x+5}=\sqrt{2x^2-4x+2+3}=\sqrt{\left(\sqrt{2}x-\sqrt{2}\right)^2+3}\)

Lại có: \(\left(\sqrt{2}x-\sqrt{2}\right)^2\ge0\forall x\)

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

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

Vậy Min y là \(2+\sqrt{3}\)

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

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

Vì \(\left(x-1\right)^2\ge0\)\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2\ge0\)\(\forall x\)\(\Rightarrow2\left(x-1\right)^2+3\ge3\)\(\forall x\)

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

\(\Rightarrow y=2+\sqrt{2\left(x-1\right)^2+3}\ge2+\sqrt{3}\)

Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(miny=2+\sqrt{3}\)\(\Leftrightarrow x=1\)

2 + \(\sqrt{2x^2-4x+5}\)

= 2 + \(\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}\)

= 2 + \(\sqrt{2\left(x^2-2x+1-1+\dfrac{5}{2}\right)}\)

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

\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=-2\)

\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)

Dấu \("="\Leftrightarrow x=-5\)

\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

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

Dấu '=' xảy ra khi x=-2

\(C=4x^2-4x+5\)

\(=4x^2-4x+1+4\)

\(=\left(2x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)