`sqrt{x^2-25}-6=3sqrt{x+5}-2sqrt{x-5}(x>=5)`

`<=>sqrt{(x-5)(x+5)}+2sqrt{x-5}=3sqrt{x+5}+6`

`<=>sqrt{x-5}(sqrt{x+5}+2)=3(sqrt{x+5}+2)`

`<=>(sqrt{x+5}+2)(sqrt{x-5}-3)=0`

Vì `sqrt{x+5}+2>0`

`<=>sqrt{x-5}-3=0`

`<=>sqrt{x-5}=3`

`<=>x-5=9<=>x=14(tm)`

Vậy `x=14`

\(\sqrt{x^2-25}-6=3\sqrt{x+5}-2\sqrt{x-5}\\ \Leftrightarrow\sqrt{\left(x-5\right)\left(x+5\right)}-6-3\sqrt{x+5}+2\sqrt{x-5}=0\\ \Leftrightarrow\left(2\sqrt{x-5}+\sqrt{\left(x-5\right)\left(x+5\right)}\right)-\left(3\sqrt{x+5}+6\right)=0\Leftrightarrow\sqrt{x-5}\left(2+\sqrt{x+5}\right)-3\left(2+\sqrt{x+5}\right)=0\\ \Leftrightarrow\left(\sqrt{x-5}-3\right)\left(2+\sqrt{x-5}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=3\\\sqrt{x-5}=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=9\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=14\)

\(\Leftrightarrow3\sqrt{x+2}-x-6+5\sqrt{x+18}-21=0\)

=>\(3\sqrt{x+2}-9+5\sqrt{x+18}-x-18=0\)

=>\(3\left(\sqrt{x+2}-3\right)+\sqrt{x+18}\left(5-\sqrt{x+18}\right)=0\)

=>\(3\cdot\dfrac{x+2-9}{\sqrt{x+2}+3}+\sqrt{x+18}\cdot\dfrac{25-x-18}{5+\sqrt{x+18}}=0\)

=>\(\left(x-7\right)\cdot\left(\dfrac{3}{\sqrt{x+2}+3}-\dfrac{\sqrt{x+18}}{5+\sqrt{x+18}}\right)=0\)

=>x-7=0

=>x=7

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)

ĐKXĐ: \(x\ge2\)

\(\dfrac{\left(\sqrt{3x-5}-\sqrt{x-2}\right)\left(\sqrt{3x-5}+\sqrt{x-2}\right)}{\sqrt{3x-5}+\sqrt{x-2}}=\dfrac{2x-3}{3}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\Rightarrow x=\dfrac{3}{2}\left(ktm\right)\\\sqrt{3x-5}+\sqrt{x-2}=3\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow\sqrt{3x-5}-2+\sqrt{x-2}-1=0\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\sqrt{3x-5}+2}+\dfrac{x-3}{\sqrt{x-2}+1}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{3}{\sqrt{3x-5}+2}+\dfrac{1}{\sqrt{x-2}+1}\right)=0\)

\(\Leftrightarrow x-3=0\)  (do \(\dfrac{3}{\sqrt{3x-5}+2}+\dfrac{1}{\sqrt{x-2}+1}>0;\forall x\ge2\))

\(\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất \(x=3\)

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)