Question

Giải phương trình:

(x+1)\(\sqrt{x^2-2x+3}\)=x²+1

(Giải bằng cách đặt ẩn phụ nha)

Answer 1

ĐK: \(x>-1\)

\(PT\Leftrightarrow a^2-\left(x+1\right)a+2x-2=0\)

\(\Leftrightarrow\left(2-a\right)\left(x-a-1\right)=0\)

.Làm nốt. 

~Ko chắc~

Answer 2

À quên: Đặt \(a=\sqrt{x^2-2x+3}\ge\sqrt{2}\)

Answer 3

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

(x+1)\(\sqrt{\left(x-1\right)^2+2}\)-(x+1)(x-1)=0

(x+1)(x-1-x+1+\(\sqrt{2}\))=0

(x+1)\(\sqrt{2}\)=0

<=>x+1=0

<=>x=-1

Answer 4

Thanks

Answer 5

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

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

(x+1)(x-1)\(\sqrt{2}\)-(\(x^2\)+1)=0

(\(x^2\)-1)\(\sqrt{2}\)-(\(x^2\)-1)-2=0

(\(x^2\)-1)(\(\sqrt{2}\)-1)=2