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

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

\(\Leftrightarrow\left(x^2+2x+1\right)^2=\left(x\sqrt[]{x^2+1}+2\sqrt[]{x^2+1}\right)^2\)

\(\Leftrightarrow x^4+4x^2+1+4x^3+2x^2+4x=x^2\left(x^2+1\right)+4\left(x^2+1\right)+4x\left(x^2+1\right)\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1=x^4+x^2+4x^2+4+4x^3+4\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1=x^4+4x^3+5x^2+4x+4\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x=\pm\sqrt[]{3}\left(Tm.x\ge-2\right)\)

Vậy nghiệm của phương trình \(\left(1\right)\) là \(x=\pm\sqrt[]{3}\)

2) \(P=\sqrt[]{x^2-2x+13}+4\sqrt[]{x-3}\)

Ta có : 

\(\sqrt[]{x^2-2x+13}=\sqrt[]{x^2-2x+1+12}=\sqrt[]{\left(x-1\right)^2+12}\ge\sqrt[]{12}=2\sqrt[]{3},\forall x\in R\)

\(4\sqrt[]{x-3}\ge0,\forall x\ge3\)

\(\Rightarrow P=\sqrt[]{x^2-2x+13}+4\sqrt[]{x-3}\ge\sqrt[]{4+12}+0=4\left(khi.x=3\right),\forall x\ge3\)

Vậy \(Min\left(P\right)=4\left(tại.x=3\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

Vậy...

b)Đk: \(x\ge-1\)

Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)

Vậy...

\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)

b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)

a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

✱ giải pt:

a.\(\sqrt{x^2-4x+4}\)\(=5\)

⇔\(\sqrt{\left(x-2\right)^2}=5\)

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

vậy....

b.\(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)

⇔ \(4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

⇔ \(4\sqrt{x+1}=16\)

⇔ \(\sqrt{x+1}=16\)

⇒ \(x+1=256\)

⇔ \(x=255\)

vậy.....

a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\)

\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)

=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi x=1

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

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

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

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

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Bài 1:

ĐKĐB suy ra $x(x+1)+y(y+1)=3x^2+xy-4x+2y+2$

$\Leftrightarrow 2x^2+x(y-5)+(y-y^2+2)=0$

Coi đây là PT bậc 2 ẩn $x$

$\Delta=(y-5)^2-4(y-y^2+2)=(3y-3)^2$Do đó:

$x=\frac{y+1}{2}$ hoặc $x=2-y$. Thay vào một trong 2 phương trình ban đầu ta thu được:

$(x,y)=(\frac{-4}{5}, \frac{-13}{5}); (1,1)$

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

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

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

\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

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

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)

Làm nốt

Điều kiện xác định: \(x\ge0;x\ne9\)

1/ \(P=\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-3}{3-\sqrt{x}}-\dfrac{3\left(3\sqrt{x}-5\right)}{x-2\sqrt{x}-3}\)

\(=\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}+\dfrac{2\sqrt{x}-3}{\sqrt{x}-3}-\dfrac{9\sqrt{x}-15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x-7\sqrt{x}-6+2x-\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(5\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{5\sqrt{x}-2}{\sqrt{x}+1}\)

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

Ta có \(P=\dfrac{5\left(\sqrt{3}+1\right)-2}{\sqrt{3}+1+1}=\dfrac{5\sqrt{3}+3}{\sqrt{3}+2}\)

c) \(P=\dfrac{5\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-7}{\sqrt{x}+1}=5-\dfrac{7}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow P\ge5-\dfrac{7}{1}=-2\)

Dấu = xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy \(P_{min}=-2\) đạt được khi \(x=0\)

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