a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

hãy thử lại với x = - 0,5

\(\dfrac{x}{x-2}+\dfrac{x+2}{x}>2\)      \(\left(ĐK:x\ne0;x\ne2\right)\)

⇔ \(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}>2\)

⇔ \(\dfrac{x^2+x^2-4}{x\left(x-2\right)}>2\)

⇔ \(\dfrac{2x^2-4}{x^2-2x}-2>0\)

⇔ \(2x^2-4-2x^2+4x>0\)

\(\Leftrightarrow4\left(x-1\right)>0\)

⇔ \(x-1>0\)

⇔ \(x>1\)

Kết hợp ĐK, ta có \(x>1\) và \(x\ne2\)

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

= \(10-2\sqrt{21}+14\sqrt{21}\)

= \(10+12\sqrt{21}\)

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

= \(-33\sqrt{2}\)

b) \(\sqrt{\text{9x}}-7\sqrt{x}=8-6\sqrt{x}\)   (ĐK: x≥0)

⇔ \(\text{3}\sqrt{\text{x}}-7\sqrt{x}+\text{6}\sqrt{x}=8\)

⇔ \(\sqrt{x}=4\)

⇔ \(x=16\)    (TMĐK)

Vậy x = 16

a) ⇔ |2x+3| = 8
⇒ \(\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x=5\\2x=-11\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

Vậy...

tử số của phân số cuối là \(\sqrt{x+1}\) hay \(\sqrt{x}+1\) thế bạn