3:
a: =>\(5\sqrt{x+1}+2\sqrt{x+1}-3\sqrt{x+1}=2\)
=>4*căn x+1=2
=>căn x+1=1/2
=>x+1=1/4
=>x=-3/4
b: \(\Leftrightarrow3\sqrt{x+2}+\sqrt{x+2}+\sqrt{x+2}=25\)
=>5*căn x+2=25
=>căn x+2=5
=>x+2=25
=>x=23
c: =>x>=-1 và x^2+2x-4=x^2+2x+1
=>-4=1(loại)
d: =>x>=3/2 và 4x^2-5x+7=4x^2-12x+9
=>x>=3/2 và 7x=2
=>x=2/7(loại)
a/
\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{8\sqrt{x}}{x-1}\\ =\left(\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{8\sqrt{x}}{x-1}\\ =\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}.\dfrac{x-1}{8\sqrt{x}}\\ =\dfrac{-4\sqrt{x}.\left(x-1\right)}{\left(x-1\right).8\sqrt{x}}=\dfrac{-4\sqrt{x}}{8\sqrt{x}}=-\dfrac{1}{2}\)
b/
\(\left(\dfrac{x+3\sqrt{x}}{\sqrt{x}+3}-2\right)\left(\dfrac{x-1}{\sqrt{x}-1}+1\right)\\ =\left(\dfrac{x+3\sqrt{x}}{\sqrt{x}+3}-\dfrac{2\sqrt{x}+6}{\sqrt{x}+3}\right)\left(\dfrac{x-1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\right)\\ =\left(\dfrac{x+3\sqrt{x}-2\sqrt{x}-6}{\sqrt{x}+3}\right)\left(\dfrac{x-1+\sqrt{x}-1}{\sqrt{x}-1}\right)\\ =\left(\dfrac{x+\sqrt{x}-6}{\sqrt{x}+3}\right)\left(\dfrac{x+\sqrt{x}-2}{\sqrt{x}-1}\right)\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\\ =\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=x-4\)
