Question

giải hộ mk nha: giải phương trình vô tỉ:

\(A=\sqrt{X^2+4X+3}=x-2\)

\(B=\sqrt{4X+20}-3\sqrt{5+X}+\frac{4}{3}\sqrt{9X+45}=6\)

Answer 1

A, đk tự tìm

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

\(\Leftrightarrow x^2+4x+3-x^2+4x-4=0\)

\(\Leftrightarrow8x-1=0\)

\(\Leftrightarrow x=\frac{1}{8}\)

B, đk tự tìm

\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9\left(x+5\right)}\)=6

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\)