a)

\(\sqrt{x}=4\Rightarrow x=4^2=16\)

c) \(x\in\varnothing\)

e)  \(\sqrt{x}=6,25\Rightarrow x=\left(6,25\right)^2=39,0625\)

b) \(\sqrt{x}=\sqrt{7}\Rightarrow x=7\)

d) \(\sqrt{x}=0\Rightarrow x=0\)

Cách đánh đề độc lạ ghê:v

a: =>x=16

b: =>x=7

c: =>x thuộc rỗng

d: =>x=0

e: =>x=(25/4)^2=625/16

\(\sqrt{x}=3\Rightarrow x=9\)

\(\sqrt{x}=\sqrt{5}\Rightarrow x=5\)

\(\sqrt{x}=0\Rightarrow x=0\)

\(\sqrt{x}=-2\Rightarrow x=\varnothing\)

a)\(\sqrt{x}=3\Rightarrow x=9\)

b)\(\sqrt{x}=\sqrt{5}\Rightarrow x=5\)

c)\(\sqrt{x}=0\Rightarrow x=0\)

d)\(\sqrt{x}=-2\Rightarrow x=4\)

\(\begin{array}{l}a)\sqrt x  - 16 = 0\\\sqrt x  = 16\\x = {16^2}\\x = 256\end{array}\)

Vậy x = 256

\(\begin{array}{l}b)2\sqrt x  = 1,5\\\sqrt x  = 1,5:2\\\sqrt x  = 0.75\\x = {(0,75)^2}\\x = 0,5625\end{array}\)

Vậy x = 0,5625

\(\begin{array}{l}c)\sqrt {x + 4}  - 0,6 = 2,4\\\sqrt {x + 4}  = 2,4 + 0,6\\\sqrt {x + 4}  = 3\\x + 4 = 9\\x = 5\end{array}\)

Vậy x = 5

\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)

ĐKXĐ: `x>=0`

`a,sqrtx=21`

`=>x=21(TMĐK)`

KL...

`b,3\sqrtx=18`

`<=>sqrtx=6`

`=>x=36(TMĐK)`

KL...

`c,sqrtx <=5`

`=>x<=25` kết hợp với điều kiện có `0<=x<=25`

KL....

`d,3sqrt(2x)>9`

`<=>sqrt(2x)>3`

`=>2x>9`

`<=>x>9/2(TMĐK)`

KL...

a. \(\sqrt{x}=21\)

Vì x\(\ge\) 0 nên bình phương 2 vế ta được:

x = 21² \(\Leftrightarrow\) x = 441

Vậy x = 441

b \(3\sqrt{x}=18\) \(\Leftrightarrow\sqrt{x}=18:3\Leftrightarrow x=\sqrt{6}\) 

Vì \(x\ge0\) nên bình phương ta được:

x = 6² \(\Leftrightarrow\) x = 36

Vậy x = 36

c. \(\sqrt{x}hoặc=\sqrt{5}\) 

\(\sqrt{x}\le\sqrt{5}\) (đk x \(\le\) 0)

\(\Rightarrow x\le5\) 

Kết hợp với đk \(\Rightarrow0\le x\le5\)  

d. \(3\sqrt{2x}>9\) 

\(\Rightarrow\sqrt{2}>3\) 

\(\Rightarrow2x>9\) 

\(\Rightarrow x>\dfrac{9}{2}\) 

Kết hợp với điều kiện \(\Rightarrow x>\dfrac{9}{2}\) 

a/ x <hoac= -23/4

b/ x=2

a/ có 2xcăn6 > 2x2=4

=> 2 căn 6 > 3+1

<=> 2 căn 6 - 3 >1

b/ có 3 căn 2 > 3 

=> 3 căn 2 - 9 > -6 

=> 6 > 9- 3 căn 2

d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

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

`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)