Tìm kiếm
Nam
Gửi bởi ĐK:
$VT^2= ( \sqrt{8-x}+\sqrt{10+x})^2\leq (1+1).(8-x+10+x)=36 \leftarrow VT^2\leq 6 VP^2=(x+1)^2+6\geq 6$$VT^2= ( \sqrt{8-x}+\sqrt{10+x})^2\leq (1+1).(8-x+10+x)=36 \leftarrow VT^2\leq 6 VP^2=(x+1)^2+6\geq 6$$VT^2= ( \sqrt{8-x}+\sqrt{10+x})^2\leq (1+1).(8-x+10+x)=36 \leftarrow VT^2\leq 6 VP^2=(x+1)^2+6\geq 6$
Suy ra PT có nghiệm duy nhất x = -1
- perfectstrong yêu thích
