a: \(\Leftrightarrow\left|2x-3\right|=7\)

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

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

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

a) |x| = 4

\(\left[ {_{x =  - 4}^{x = 4}} \right.\)

Vậy \(x \in \{ 4; - 4\} \)

b) |x| = \(\sqrt 7 \)

\(\left[ {_{x =  - \sqrt 7 }^{x = \sqrt 7 }} \right.\)

Vậy \(x \in \{ \sqrt 7 ; - \sqrt 7 \} \)

c) ) |x+5| = 0

x+5 = 0

x = -5

Vậy x = -5

d) \(\left| {x - \sqrt 2 } \right|\) = 0

x - \(\sqrt 2 \) = 0

x = \(\sqrt 2 \)

Vậy x =\(\sqrt 2 \)

a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)

=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)

=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)

=>\(-2x=\dfrac{1}{4}\)

=>\(2x=-\dfrac{1}{4}\)

=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)

b: ĐKXĐ: x>=0

\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)

=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

`a)sqrt{x^2-2x+1}=2`

`<=>sqrt{(x-1)^2}=2`

`<=>|x-1|=2`

`**x-1=2<=>x=3`

`**x-1=-1<=>x=-1`.

Vậy `S={3,-1}`

`b)sqrt{x^2-1}=x`

Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)

`<=>x>=1`

`pt<=>x^2-1=x^2`

`<=>-1=0` vô lý

Vậy pt vô nghiệm

`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`

`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`

`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`

`<=>2sqrt{x-5}=4`

`<=>sqrt{x-5}=2`

`<=>x-5=4`

`<=>x=9(tmđk)`

Vậy `S={9}.`

`d)x-5sqrt{x-2}=-2(x>=2)`

`<=>x-2-5sqrt{x-2}+4=0`

Đặt `a=sqrt{x-2}`

`pt<=>a^2-5a+4=0`

`<=>a_1=1,a_2=4`

`<=>sqrt{x-2}=1,sqrt{x-2}=4`

`<=>x_1=3,x_2=18`,

`e)2x-3sqrt{2x-1}-5=0`

`<=>2x-1-3sqrt{2x-1}-4=0`

Đặt `a=sqrt{2x-1}(a>=0)`

`pt<=>a^2-3a-4=0`

`a-b+c=0`

`<=>a_1=-1(l),a_2=4(tm)`

`<=>sqrt{2x-1}=4`

`<=>2x-1=16`

`<=>x=17/2(tm)`

Vậy `S={17/2}`

d.

ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:

$a^2+2-5a=-2$

$\Leftrightarrow a^2-5a+4=0$

$\Leftrightarrow (a-1)(a-4)=0$

$\Rightarrow a=1$ hoặc $a=4$

$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$

$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)

e. ĐKXĐ: $x\geq \frac{1}{2}$

Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:

$a^2+1-3a-5=0$

$\Leftrightarrow a^2-3a-4=0$

$\Leftrightarrow (a+1)(a-4)=0$

Vì $a\geq 0$ nên $a=4$

$\Leftrightarrow \sqrt{2x-1}=4$

$\Leftrightarrow x=\frac{17}{2}$

a.

$\sqrt{x^2-2x+1}=2$

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

$\Leftrightarrow |x-1|=2$

$\Rightarrow x-1=\pm 2$

$\Leftrightarrow x=3$ hoặc $x=-1$ (đều thỏa mãn)

b. ĐKXĐ: $x\geq 1$ hoặc $x\leq -1$

PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2-1=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 1=0\end{matrix}\right.\) (vô lý)

Vậy pt vô nghiệm

c. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

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

$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)

Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
    \(3x=16+2=18\)
    \(x=18:3=6\)
    Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
    \(2x+1=5+11=16\)
    \(2x=16-1=15\)
    \(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
    \(2x-1=-5+11=6\)
    \(2x=6+1=7\)
    \(x=7:2=3,5\)
    Vậy \(x=\left\{7,5;3,5\right\}\) 
    (Câu này mình không chắc chắn lắm)   
    (Học sinh lớp 6 đang làm bài này)    

Bài 4:

a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)

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

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

b: C-6<0

=>C<6

=>\(2\sqrt{x}< 6\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)

Bài 3

a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)

Vậy PT có nghiệm x=6

b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)

+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)

+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)

Bài 4

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

\(Vậy\) \(C=2\sqrt{x}\)

Bài 2:

a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)

\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)

\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)

\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)

\(=8\sqrt{5}\)

b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(\sqrt{7}+3\right)^2}\)

\(=\sqrt{7}-2-\sqrt{7}-3\)

\(=-5\)

\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)

3: 

a: =>|x-1|=4

=>x-1=4 hoặc x-1=-4

=>x=-3 hoặc x=5

b: =>|6x-5|=4

=>6x-5=4 hoặc 6x-5=-4

=>6x=1 hoặc 6x=9

=>x=1/6 hoặc x=3/2

\(a,ĐK:x\ge3\\ PT\Leftrightarrow x-3=5\Leftrightarrow x=8\left(tm\right)\\ b,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-1=3\Leftrightarrow x=2\left(tm\right)\\ c,Vì.\sqrt{1-x}\ge0>-1.nên.pt.vô.nghiệm\\ d,PT\Leftrightarrow\left|x-1\right|=1\Leftrightarrow\left[{}\begin{matrix}x-1=1\\1-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

a) \(\sqrt{x-3}=5\) (1)

ĐKXĐ: \(x\ge3\)

\(\left(1\right)\Leftrightarrow x-3=25\)

\(\Leftrightarrow x=28\) (nhận)

Vậy \(x=28\)

b) \(\sqrt{2x-1}=\sqrt{3}\)   (2)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\left(2\right)\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\) (nhận)

Vậy \(x=2\)

c) \(\sqrt{1-x}=-1\)

Không tìm được \(x\) vì \(\sqrt{1-x}\ge0\) (với mọi \(x\le1\))

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

ĐKXĐ: Với mọi \(x\in R\)

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

\(\Leftrightarrow x-1=1\) (khi \(x\ge1\)) hoặc \(1-x=1\) (khi \(x< 1\))

* \(x-1=1\)

\(\Leftrightarrow x=2\) (nhận)

* \(1-x=1\)

\(\Leftrightarrow x=0\) (nhận)

Vậy \(x=0;x=2\)

a: \(\sqrt{\left(x+1\right)^2}=5\)(ĐKXĐ: \(x\in R\))

=>|x+1|=5

=>\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

b: Sửa đề: \(5\sqrt{9x-9}-\sqrt{4\left(x-1\right)}+\sqrt{36\left(x-1\right)}-18=0\)

ĐKXĐ: x>=1

\(PT\Leftrightarrow5\cdot3\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0\)

=>\(15\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}=18\)

=>\(19\sqrt{x-1}=18\)

=>\(\sqrt{x-1}=\dfrac{18}{19}\)

=>\(x-1=\left(\dfrac{18}{19}\right)^2=\dfrac{324}{361}\)

=>\(x=\dfrac{324}{361}+1=\dfrac{324+361}{361}=\dfrac{685}{361}\)

Lời giải:

a. PT $\Leftrightarrow |x+1|=5$

$\Leftrightarrow x+1=\pm 5\Leftrightarrow x=4$ hoặc $x=-6$

b. ** Sửa $x-9$ thành $x-1$

ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow 5\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0$

$\Leftrightarrow (5-2+6)\sqrt{x-1}=18$

$\Leftrightarrow 9\sqrt{x-1}=18$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

a, ĐKXĐ:\(x\ge1\)

\(\sqrt{x-1}=3\\ \Rightarrow x-1=9\\ \Rightarrow x=10\)

\(b,x^2-64=0\\ \Rightarrow\left(x-8\right)\left(x+8\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\\ c,x^2+16=25\\ \Rightarrow x^2=9\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ d,ĐKXĐ:x\ge0\\ \left|\sqrt{x}-3\right|+3=9\\ \Rightarrow\left|\sqrt{x}-3\right|=6\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}-3=-6\\x-3=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(vô.lí\right)\\x=9\left(tm\right)\end{matrix}\right.\)

a) \(\sqrt{x}< \sqrt{3}\Rightarrow x< 3\Rightarrow0\le x< 3\)

b) \(\sqrt{3x}< 6\Rightarrow3x< 36\Rightarrow x< 12\Rightarrow0\le x< 12\)

c) \(\dfrac{1}{2}\sqrt{5x}< 10\Rightarrow\sqrt{5x}< 20\Rightarrow5x< 400\Rightarrow x< 80\Rightarrow0\le x< 80\)

a) \(0\le x< 3\)

b) \(0\le x< 12\)