\(ĐK:x^2-3x+5\ge0\)

Đặt \(\sqrt{x^2-3x+5}=a\ge0\)

\(PT\Leftrightarrow a+a^2-5=7\\ \Leftrightarrow a^2+a-12=0\\ \Leftrightarrow\left(a-3\right)\left(a+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow\sqrt{x^2-3x+5}=3\\ \Leftrightarrow x^2-3x+5=9\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

đặt \(x^2-3x=y\)

\(pt\Leftrightarrow\sqrt{y+5}+y=7\\ \Leftrightarrow\sqrt{y+5}=7-y\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=\left(7-y\right)^2\\7-y\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=49-14y+y^2\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-15y+44=0\\y\le7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(y^2-11y\right)-\left(4y-44\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-11\right)\left(y-4\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=4\\y=11\end{matrix}\right.\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x-4=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+1\right)\\y\le7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\\y\le7\end{matrix}\right.\)

Vậy \(x\in\left\{4;-1\right\}\)

TUi lớp 7

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:

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

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

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

`<=>\sqrt{3x}=1`

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

`<=>\sqrt{x+1}=3`

`<=>x+1=9`

`<=>x=8` (t/m)

ĐKXĐ: \(x\in R\)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

=>\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x-4=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x+1-5=0\)

=>\(\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+14}-3+\left(x+1\right)^2=0\)

=>\(\dfrac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+14-9}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>

\(\dfrac{3x^2+6x+3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+5}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\dfrac{3\left(x^2+2x+1\right)}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x^2+2x+1\right)}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5}{\sqrt{5x^2+10x+14}+3}+1\right)=0\)

=>\(\left(x+1\right)^2=0\)

=>x+1=0

=>x=-1(nhận)