Question

\(\sqrt{a+1}-4a^2=\sqrt{3a}-1\\ giảidùmmìnhvớisángmìnhvừathithử\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}3a>=0\\a+1>=0\end{matrix}\right.\)

=>a>=0

\(\sqrt{a+1}-4a^2=\sqrt{3a}-1\)

=>\(\sqrt{a+1}-\sqrt{3a}=4a^2-1\)

=>\(\dfrac{a+1-3a}{\sqrt{a+1}+\sqrt{3a}}=\left(2a-1\right)\left(2a+1\right)\)

\(\Leftrightarrow\dfrac{-2a+1}{\sqrt{a+1}+\sqrt{3a}}-\left(2a-1\right)\left(2a+1\right)=0\)

=>\(\dfrac{2a-1}{\sqrt{a+1}+\sqrt{3a}}+\left(2a-1\right)\left(2a+1\right)=0\)

=>\(\left(2a-1\right)\cdot\left(\dfrac{1}{\sqrt{a+1}+\sqrt{3a}}+2a+1\right)=0\)

=>2a-1=0

=>a=1/2(nhận)