\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\)

=>(x-3)(2x+5-x-3)=0

=>(x-3)(x+2)=0

=>x=3 hoặc x=-2

     x²-9=(x-3)(2x-5)
(=) (x-3)(x+3)=(x-3)(2x-5)

(=) (x-3)(x+3)-(x-3)(2x-5)=0

(=) (x-3)(x+3-2x+5)=0
(=) (x-3)(8-x)=0

(=)x-3=0 hoặc 8-x=0

(=)x=0 hoặc x=8

Vậy S=\(\left\{0;8\right\}\)

=>\(\left(\dfrac{x^2-8}{2008}-1\right)+\left(\dfrac{x^2-7}{2009}-1\right)=\left(\dfrac{x^2-6}{2010}-1\right)+\left(\dfrac{x^2-5}{2011}-1\right)\)

=>x^2-2016=0

=>x^2=2016

=>\(x=\pm\sqrt{2016}\)

\(đk:4x^2-12x+44\ge0\left(luôn-đúng\right)\)

\(x^2+\sqrt{4x^2-12x+44}=3x+4\)

\(\Leftrightarrow x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

\(\Leftrightarrow x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

\(đặt:\sqrt{x^2-3x+11}=t\left(t\ge0\right)\)

\(\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-3x+11}=3\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

ĐKXĐ: ...

\(x^2+\sqrt{4x^2-12x+44}=3x+4\)

\(\Leftrightarrow\sqrt{4x^2-12x+44}=3x+4-x^2\)

\(\Leftrightarrow4x^2-12x+44=\left(3x+4-x^2\right)^2\)

\(\Leftrightarrow4x^2-12x+44=x^4-6x^3+x^2+24x+16\)

\(\Leftrightarrow x^4-6x^3-3x^2+36x-28=0\)

...........

\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

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

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

\(\Leftrightarrow x-1-\sqrt{x-1}-1=0\) (1)

Đặt \(\sqrt{x-1}\) = t (t \(\ge0\))

pttt : t² - t - 1 =0

\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\) 

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1-\sqrt{5}}{2}\left(ktm\right)\\t=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)

=> \(\sqrt{x-1}=\dfrac{1+\sqrt{5}}{2}\)

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

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

p/s: thử lại hộ mình nhaa

Hummmm

Dạ em không biết ạ,tại vì em mới học lớp 4 ạ,em xin lỗi ạ

ĐKXĐ: \(x\ge1\)

Do \(\sqrt{x-\sqrt{x^2-1}}.\sqrt{x+\sqrt{x^2-1}}=\sqrt{x^2-x^2+1}=1\)

Đặt \(\sqrt{x-\sqrt{x^2-1}}=t\Rightarrow\sqrt{x+\sqrt{x^2-1}}=\dfrac{1}{t}\)

Phương trình trở thành:

\(t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\Rightarrow t=1\)

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

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

\(\Rightarrow x^2-2x+1=x^2-1\)

\(\Rightarrow x=1\) (thỏa mãn)

ĐKXĐ: \(x\ge1\)

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

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

\(\Leftrightarrow\dfrac{\sqrt{x-1}.x}{\sqrt{x+1}-1}=x\sqrt{x}\)

\(\Leftrightarrow\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}=\sqrt{x}\)

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

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

\(\Leftrightarrow2x-1+2\sqrt{x^2-x}=x^2+x\)

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

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

\(\Leftrightarrow x^2-x-1=0\)