\(\sqrt{3x^2+33}+3\sqrt{x}=2x+7\)(ĐKXĐ: x>=0)
=>\(\sqrt{3x^2+33}-6+3\sqrt{x}-3=2x-2\)
=>\(\dfrac{3x^2+33-36}{\sqrt{3x^2+33}+6}+3\left(\sqrt{x}-1\right)=2\left(x-1\right)\)
=>\(\dfrac{3x^2-3}{\sqrt{3x^2+33}+6}+3\left(\sqrt{x}-1\right)-2\left(x-1\right)=0\)
=>\(\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+1\right)}{\sqrt{3x^2+33}+6}+3\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
=>\(\left(\sqrt{x}-1\right)\left(\dfrac{3\left(\sqrt{x}+1\right)\left(x+1\right)}{\sqrt{3x^2+33}+6}+3-2\left(\sqrt{x}+1\right)\right)=0\)
=>\(\sqrt{x}-1=0\)
=>x=1(nhận)
Đúng 1
Bình luận (0)