Question

Phương pháp 6. Biến đổi về dạng \(A^2=B^2\)

a \(x^2+4\sqrt{x+3}=3x+6\)

b \(4x^2+14x+11=4\sqrt{6x+10}\)

c \(4\sqrt{x+1}=x^2-5x+14\)

Answer 1

a) ĐKXĐ : \(x\ge-3\)\(pt\Leftrightarrow x^2-2x+1=x+3-4\sqrt{x+3}+4\Leftrightarrow\left(x-1\right)^2=\left(\sqrt{x+3}-2\right)^2\Leftrightarrow x-1=\sqrt{x+3}-2\Leftrightarrow x+1=\sqrt{x+3}\Leftrightarrow\left(x+1\right)^2=x+3\left(x\ge-1\right)\Leftrightarrow x^2+2x+1=x+3\Leftrightarrow x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(tmdk\right)\\x=-2\left(kotm\right)\end{matrix}\right.\)