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

\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) 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 \(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)

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

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

\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)

b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)

\(2\sqrt{x}-1>0\);\(4x>0\)

\(\Rightarrow x>0\)thì \(A>A^2\)

câu c là \(\left|A\right|>\frac{1}{4}\)

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

a 

ĐK: \(x^2-2x+1>0\)

PT \(\Leftrightarrow\sqrt{\left(x-1\right)^2}+x-6x+9=0\)

\(\Leftrightarrow\left|x-1\right|-5x+9=0\\ \Leftrightarrow\left|x-1\right|=-9+5x\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-9+5x\\1-x=-9+5x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=\dfrac{10}{6}\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}2x^2-3>0\\4x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>\dfrac{\sqrt{6}}{2}\\x< -\dfrac{\sqrt{6}}{2}\end{matrix}\right.\\x>\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x>\dfrac{\sqrt{6}}{2}\)

PT \(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x^2-4x=0\\ \Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(\left\{{}\begin{matrix}1-x^2\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x=1\)

PT \(\Leftrightarrow1-x^2=x-1\)

\(\Leftrightarrow x^2+x-2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

a: ĐKXĐ: x>=-3/2

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

=>\(x^2+4=2x+3\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1(nhận)

b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))

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

=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>x=4/3(nhận) hoặc x=-2(loại)

c:

Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)

ĐKXĐ: \(x>=-3\)

\(\sqrt{4x+12}=\sqrt{9x+27}-5\)

=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)

=>\(-\sqrt{x+3}=-5\)

=>x+3=25

=>x=22(nhận)

d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)

=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)

=>\(4x^2-6x+1=4x^2-20x+25\)

=>\(-6x+20x=25-1\)

=>\(14x=24\)

=>x=12/7(nhận)

a, ĐKXĐ: \(x^2-4x+4\ge0\Rightarrow\left(x-2\right)^2\ge0\left(luônđúng\right)\)

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

b,\(ĐKXĐ:1-4x+4x^2\ge0\Rightarrow\left(1-2x\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{1-4x+4x^2}=5\\ \Rightarrow\left|1-2x\right|=5\\ \Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

d, ĐKXĐ: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)

\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

c, ĐKXĐ: \(1-2x+x^2\ge0\Rightarrow\left(1-x\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{1-2x+x^2}-6=0\\ \Rightarrow\left|1-x\right|=6\\ \Rightarrow\left[{}\begin{matrix}1-x=-6\\1-x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

e, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)

\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)

a) \(\sqrt{x^2-4x+4}=1\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)

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

b) \(\sqrt{1-4x+4x^2}=5\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

c) \(\sqrt{x\left(1-2x+x^2\right)}-6=0\)

\(\Leftrightarrow\left(\sqrt{x\left(1-x\right)^2}\right)^2=36\Leftrightarrow x\left(1-x\right)^2=36\)

\(\Leftrightarrow x-2x^2+x^3-36=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+2x+9\right)=0\)

\(\Leftrightarrow x=4\)(do \(x^2+2x+9=\left(x+1\right)^2+8>0\))

d) \(\sqrt{9x^2}=2x+1\)

\(\Leftrightarrow3\left|x\right|=2x+1\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\\-3x=2x+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)

e) \(\sqrt{9-6x+x^2}=x\left(1\right)\left(đk:x\ge0\right)\)

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

TH1: \(0\le x\le3\)

\(\left(1\right)\Leftrightarrow3-x=x\Leftrightarrow x=\dfrac{3}{2}\)

TH2: \(x>3\)

\(\left(1\right)\Leftrightarrow x-3=x\Leftrightarrow-3=0\left(vn\right)\)