\(a,\sqrt{9x^2}=2x+1\\ < =>\sqrt{\left(3x\right)^2}=2x+1\\ < =>\left|3x\right|=2x+1\)
Ta có:\(\left|3x\right|=3xkhi3x\ge0\)
\(\left|3x\right|=-3xkhi3x< 0\)
(*) 3x\(\ge\)0, ta có phương trình:
\(3x=3x+1\\ < =>3x-2x=1\\ < =>x=1\)
(*)3x<1, ta có phương trình:
\(-3x=2x+1\\ < =>-3x-2x=1\\ < =>-5x=1\\ < =>x=-\dfrac{1}{5}\)
\(b,\sqrt{x^4}=7\\ < =>\sqrt{\left(x^2\right)^2}=7\\ < =>\left|x^2\right|=7\\ < =>x^2=7\\ < =>x=\pm\sqrt{7}\)