Question

Giúp mình từ câu 15 đến 18 với

Answer 1

15)

ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{x+3}+\sqrt{x-1}=2\)

\(\Leftrightarrow2x+2+2\sqrt{\left(x+3\right)\left(x-1\right)}=4\)

\(\Leftrightarrow2\sqrt{\left(x+3\right)\left(x-1\right)}=2-2x\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-1\right)}=1-x\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=\left(1-x\right)^2\)

\(\Leftrightarrow x^2-x+3x-3=x^2-2x+1\)

\(\Leftrightarrow2x-3+2x-1=0\)

\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow4x=4\)

hay x=1(thỏa ĐK)

Vậy: S={1}