\(a.11+2\sqrt{30}=6+2.\sqrt{6}.\sqrt{5}+5=\left(\sqrt{6}+\sqrt{5}\right)^2\)

\(b.10+2\sqrt{21}=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)

\(c.6x-\sqrt{x}-1=6x+2\sqrt{x}-3\sqrt{x}-1=2\sqrt{x}\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}+1\right)=\left(3\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\) \(d.14-2\sqrt{45}=14-6\sqrt{5}=9-2.3\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)

\(e.4x-3\sqrt{x}-1=4x-4\sqrt{x}+\sqrt{x}-1=4\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(4\sqrt{x}+1\right)\) \(j.12-2\sqrt{27}=9-2.3.\sqrt{3}+3=\left(3-\sqrt{3}\right)^2\)

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

a: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>2x=-2 hoặc 2x=4

=>x=2 hoặc x=-1

c: \(\Leftrightarrow\left|x-3\right|=11-x\)

=>x<=11 và (x-3)^2=(11-x)^2

=>x<=11 và x^2-6x+9=x^2-22x+121

=>x<=11 và 16x=112

=>x=7

d:

ĐKXĐ: 3x+19>=0

=>x>=-19/3

PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9

=>x>=-3 và x^2+6x+9-3x-19=0 

=>x>=-3 và (x+5)(x-2)=0

=>x=2

e: =>\(\sqrt{x^2+x+5}=x+1\)

=>x>=-1 và x^2+x+5=x^2+2x+1

=>x>=-1 và 2x+1=x+5

=>x=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. 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$

$\Leftrightarrow \sqrt{2x}=3$

$\Leftrightarrow 2x=9$

$\Leftrightarrow x=\frac{9}{2}$ (tm)

b.

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$

$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$

$\Leftrightarrow 3\sqrt{x+2}=15$

$\Leftrightarrow \sqrt{x+2}=5$

$\Leftrightarrow x+2=25$

$\Leftrightarrow x=23$ (tm)

c.

$\sqrt{(x-2)^2}=12$

$\Leftrightarrow |x-2|=12$

$\Leftrightarrow x-2=12$ hoặc $x-2=-12$

$\Leftrightarrow x=14$ hoặc $x=-10$

e.

PT $\Leftrightarrow |2x-1|-x=3$

Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

f.

ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$

$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$

$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$

$\Leftrightarrow x=2$ hoặc $x=5$ (tm)

h. ĐKXĐ: $x\leq \frac{3}{2}$

PT $\Leftrightarrow \sqrt{3-2x}=x+2$

\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)

\(\Leftrightarrow x=-3+2\sqrt{2}\) (tm)

Vậy.......

a,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) (*)(đk \(x\ge-2\))

<=> \(\sqrt{\left(x+2\right)-4\sqrt{x+2}+4}+\sqrt{\left(x+2\right)-6\sqrt{x+2}+9}\)=1

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

<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|\)=1 (1)

TH1: \(0\le\sqrt{x+2}< 2\)

Từ (1) =>\(2-\sqrt{x+2}+3-\sqrt{x+2}=1\)

<=> \(5-2\sqrt{x+2}=1\) <=> \(2\sqrt{x+1}=4\) <=> \(\sqrt{x+1}=2\)

<=> \(x+1=4\) <=> x=3(không t/m \(\sqrt{x+2}\le2\))

TH2 : \(2\le\sqrt{x+2}\le3\)

Từ (1) =>\(\sqrt{x+2}-2+3-\sqrt{x+2}=1\)

<=> \(1=1\) (luôn đúng)

Từ TH2 <=> 4\(\le x+2\le9\) <=> \(2\le x\le7\)

TH3 \(\sqrt{x+2}>3\)

Từ (1) => \(\sqrt{x+2}-2+\sqrt{x+2}-3=1\)

<=> \(2\sqrt{x+2}=6\) <=> \(\sqrt{x+2}=3\) <=> \(x+2=9\) <=> x=7 (không t/m \(\sqrt{x+2}>3\))

Vậy pt (*) có tập nghiệm S=\(\left\{2\le x\le7\right\}\)

b, \(x^2-10x+27=\sqrt{6-x}+\sqrt{x-4}\) (*) (đk :\(4\le x\le6\))

Vs a,b \(\ge0\) ta có \(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a^2+b^2\right)}\)(tự CM nha)

Dấu "=" xảy ra <=> a=b

Áp dụng bđt trên ta có: \(\sqrt{6-x}+\sqrt{x-4}\le\sqrt{2\left(6-x+x-4\right)}=\sqrt{2.2}=2\)

<=> \(\sqrt{6-x}+\sqrt{x-4}\le2\)(1)

Lại có: \(x^2-10x+27=x^2-10x+25+2=\left(x-5\right)^2+2\ge2\)

<=> \(x^2-10x+27\ge2\) (2)

Từ (1),(2) => Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}6-x=x-4\\x-5=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}6+4=2x\\x=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\left(tm\right)\)

Vậy pt (*) có tập nghiệm S=\(\left\{5\right\}\)

c, \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)(*) (đk: x\(\ge0\))

<=> \(x\left(x-2\right)-\sqrt{x}\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)

<=> \(\left(x-\sqrt{x}\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)

<=> \(\sqrt{x}\left(\sqrt{x}-1\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)

<=> \(\left(\sqrt{x}-1\right)\left[\sqrt{x}\left(x-2\right)-4\right]=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}\left(x-2\right)-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}\left(x-2\right)=4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x\left(x-2\right)^2=16\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x\left(x^2-4x+4\right)-16=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=1\\x^3-4x^2+4x-16=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=1\\x^2\left(x-4\right)+4\left(x-4\right)=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\\left(x^2+4\right)\left(x-4\right)=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\left(tm\right)\)

Vậy pt (*) có tập nghiệm S=\(\left\{1;4\right\}\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

a) Đk: \(\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)

\(\sqrt{x^2-1}-x^2+1=0\)

\(\Leftrightarrow x^2-1-\sqrt{x^2-1}= 0\)

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

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

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

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

\(\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)\)

\(\left(2\right)\Leftrightarrow x=\pm1\left(N\right)\)

Kl: \(x=\pm\sqrt{2}\), \(x=\pm1\)

b) Đk: \(\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)

\(\sqrt{x^2-4}-x+2=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.\)

kl: x=2

c) \(\sqrt{x^4-8x^2+16}=2-x\)

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

\(\Leftrightarrow\left|x^2-4\right|=2-x\) (*)

Th1: \(x^2-4< 0\Leftrightarrow-2< x< 2\)

(*) \(\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.\)

Th2: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)

(*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)

Kl: x=-3, x=-1,x=2

d) \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)

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

\(\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}\) (*)

Th1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)

(*) \(\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)\)

Th2: \(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)

(*) \(\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)\)

Kl: \(x=\dfrac{2-\sqrt{2}}{3}\), \(x=\dfrac{-4+\sqrt{2}}{3}\)

e) Đk: \(x\ge-\dfrac{3}{2}\)

\(\sqrt{4^2-9}=2\sqrt{2x+3}\) \(\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}\) \(\Leftrightarrow7=8x+12\)

\(\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)\)

kl: \(x=-\dfrac{5}{8}\)

f) Đk: x >/ 5

\(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\left(N\right)\)

kl: x=9