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

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

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

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

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

ĐKXĐ: \(x\ge0\)

\(\dfrac{-3\sqrt{x}-5}{\sqrt{x}+1}=0\)

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

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

\(\Rightarrow\) Không tồn tại x thỏa mãn

\(a.\)

\(x-3\sqrt{x}=0\)

\(\Rightarrow\left(\sqrt{x}\right)^2-3\sqrt{x}=0\)

\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-3\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\\sqrt{x}-3=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\\sqrt{x}=3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

Vậy : \(x\in\left\{0;9\right\}\)

`#3107.101107`

`1/2x + 4/5 = 2x - 8/5`

`=> 1/2x - 2x = -4/5 - 8/5`

`=> -3/2x = -12/5`

`=> x = -12/5 \div (-3/2)`

`=> x = 8/5`

Vậy, `x = 8/5`

_____

`\sqrt{x} = 5`

`=> x = 5^2`

`=> x = 25`

Vậy, `x = 25`

___

`x^2 = 3`

`=> x^2 =  (+-\sqrt{3})^2`

`=> x = +- \sqrt{3}`

Vậy, `x \in {-\sqrt{3}; \sqrt{3}}.`

\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\\x=25\end{matrix}\right.\)

a)\(\sqrt{x^2+x+\frac{1}{4}}-\sqrt{4-2\sqrt{3}}=0\)

\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=0\)

\(\Leftrightarrow x+\frac{1}{2}-\sqrt{3}+1=0\)

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

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

b)\(x-5\sqrt{x}+6=0\)

\(\Leftrightarrow x-2\sqrt{x}-3\sqrt{x}+6=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}-2=0\\\sqrt{x}-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=2\\\sqrt{x}=3\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=9\end{array}\right.\)

đk x>=0

<=>\(x=\sqrt[5]{x}\)

<=>\(x^5=x\) 

<=>x(x^4-1)=0

<=>\(\orbr{\begin{cases}x=0\\x^4=1\end{cases}}\)

1) \(x-2\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

2) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

3) \(2x+5\sqrt{x}=0\Rightarrow\sqrt{x}\left(2\sqrt{x}+5\right)=0\Rightarrow\sqrt{x}=0\)(Vì \(\sqrt{x}\ge0\Rightarrow2\sqrt{x}+5>0\))\(\Rightarrow x=0\)

Điều kiện xác định : \(x\ge1\)

\(x+5-5\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)-5\sqrt{x-1}+6=0\)

Đặt \(t=\sqrt{x-1},t\ge0\)pt trên trở thành \(t^2-5t+6=0\Leftrightarrow\left(t-2\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=2\\t=3\end{cases}}\)

Từ đó tìm được các giá trị của x

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)