Question

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

Answer 1

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

\(\Leftrightarrow4x^2+4x+1=x^2-x-2+2\sqrt{x^2-x-2}+1\)

\(\Leftrightarrow4x+4x+1-x^2+x+2-1=2\sqrt{x^2-x-2}+1\)

\(\Leftrightarrow3x^2+5x-2=2\sqrt{x^2-x-2}\)

\(\Leftrightarrow\int^{3x^2+5x-2=0}_{4\left(x^2-x-2\right)=3x^2+5x-2}\)..............

Answer 2

Đặt \(y=\sqrt{x^2-x-2}\left(y\ge0\right)\)rồi tính nha

S=\(\frac{1-\sqrt{13}}{2};\frac{1+\sqrt{13}}{2};3;-2\)

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