\(2.< =>5\sqrt{x-1}-6\sqrt{x-1}-3\sqrt{x-1}=2\sqrt{2x-3}\)

\(< =>\sqrt{x-1}\left(5-6+3\right)=2\sqrt{2x-3}\)

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

\(< =>x-1=2x-3\)

\(< =>x=2\)

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.......

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

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

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

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

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

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

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

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

a) \(4\sqrt{2x+1}-\sqrt{8x+4}+\dfrac{1}{2}\sqrt{32x+16}=12\) (ĐK: \(x\ge-\dfrac{1}{2}\)) 

\(\Leftrightarrow4\sqrt{2x+1}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{2}\cdot4\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}-2\sqrt{2x+1}+2\sqrt{2x+1}=12\)

\(\Leftrightarrow4\sqrt{2x+1}=12\)

\(\Leftrightarrow\sqrt{2x+1}=\dfrac{12}{4}\)

\(\Leftrightarrow2x+1=3^2\)

\(\Leftrightarrow2x=9-1\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=\dfrac{8}{2}\)

\(\Leftrightarrow x=4\left(tm\right)\)

b) \(\sqrt{4x^2-4x+1}=5\)

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

\(\Leftrightarrow\left|2x-1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{2}\\x=-\dfrac{4}{2}\end{matrix}\right.\)

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

c) \(\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}=-\dfrac{1}{2}\)(ĐK: \(x\ge0;x\ne1\))

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

\(\Leftrightarrow-\sqrt{x}+1=4\sqrt{x}-6\)

\(\Leftrightarrow4\sqrt{x}+\sqrt{x}=1+6\)

\(\Leftrightarrow5\sqrt{x}=7\)

\(\Leftrightarrow\sqrt{x}=\dfrac{7}{5}\)

\(\Leftrightarrow x=\dfrac{49}{25}\left(tm\right)\)

1.ĐK: \(x\ge\dfrac{1}{4}\)

bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)

\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)

\(\Leftrightarrow20x^2-x-1>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)

2.ĐK: \(-2\le x\le\dfrac{5}{2}\)

bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)

\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)

\(\Leftrightarrow x^2< -x^2+x+6\)

\(\Leftrightarrow-2x^2+x+6>0\)

\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)

3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)

.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)

\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)

\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)

*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)

*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

a. \(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\)      ĐK: \(x\ge0\)

<=> \(\sqrt{9.2x}+2\sqrt{4.2x}-3\sqrt{2x}=12\)

<=> \(3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

<=> \(\sqrt{2x}\left(3+4-3\right)=12\)

<=> \(4\sqrt{2x}=12\)

<=> \(\sqrt{2x}=12:4\)

<=> \(\sqrt{2x}=3\)

<=> 2x = 3²

<=> 2x = 9

<=> \(x=\dfrac{9}{2}\) (TM)

b. \(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\)          ĐK: \(x\ge-2\)

<=> \(\sqrt{9\left(x+2\right)}+2\sqrt{36\left(x+2\right)}-\sqrt{4\left(x+2\right)}=26\)

<=> \(3\sqrt{x+2}+72\sqrt{x+2}-2\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}\left(3+72-2\right)=26\)

<=> \(73\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}=\dfrac{26}{73}\)

<=> x + 2 = \(\left(\dfrac{26}{73}\right)^2\)

<=> x + 2 = \(\dfrac{676}{5329}\)

<=> \(x=\dfrac{676}{5329}-2\)

<=> \(x=-1,873146932\) (TM)

c. \(\sqrt{\left(x-2\right)^2}=10\)

<=> \(\left|x-2\right|=10\)

<=> \(\left[{}\begin{matrix}x-2=10\left(x\ge2\right)\\x-2=-10\left(x< 2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=12\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)

d. \(\sqrt{9x^2-6x+1}=15\)

<=> \(\sqrt{\left(3x-1\right)^2}=15\)

<=> \(\left|3x-1\right|=15\)

<=> \(\left[{}\begin{matrix}3x-1=15\left(x\ge\dfrac{16}{3}\right)\\3x-1=-15\left(x< \dfrac{16}{3}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{16}{3}\left(TM\right)\\x=\dfrac{-14}{3}\left(TM\right)\end{matrix}\right.\)

e. \(\sqrt{3x+4}=3x-8\)        ĐK: \(x\ge\dfrac{-4}{3}\)

<=> 3x + 4 = (3x - 8)²

<=> 3x + 4 = 9x² - 48x + 64

<=> 9x² - 3x - 48x + 64 - 4 = 0

<=> 9x² - 51x + 60 = 0

<=> 9x² - 36x - 15x + 60 = 0

<=> 9x(x - 4) - 15(x - 4) = 0

<=> (9x - 15)(x - 4) = 0

<=> \(\left[{}\begin{matrix}9x-15=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{15}{9}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)

b) Ta có pt \(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

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

...

a) Đặt \(\sqrt{x^2-4x-5}=a\left(a\ge0\right)\)

Ta có pt \(\Leftrightarrow2a^2-3a-2=0\Leftrightarrow\left(a-2\right)\left(2a+1\right)=0\)

...

mình không hiểu gì hết bạn ơi

Câu 1:

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

- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm

- Nhận thấy \(x=-1\) là 1 nghiệm

- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:

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

\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)

\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)

\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)

\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)

Vậy pt có nghiệm \(x=\pm1\)

Câu 2:

ĐKXĐ: \(x\ge1\)

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

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

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

- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:

\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)

- Nếu \(1\le x< 2\) pt trở thành:

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

Vậy nghiệm của pt là \(x\ge2\)

Câu 3:

Bình phương 2 vế ta được:

\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)

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

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

Đặt \(x^2+x+1=a>0\) pt trở thành:

\(a\left(a+3\right)=4\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

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

Câu 5:

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

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

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

\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)

Vậy nghiệm của pt là \(5\le x\le10\)