Ta có: \(\sqrt{x^2-3x}=-x^2+3x+6\)
\(\Leftrightarrow\left(\sqrt{x^2-3x}\right)^2=\left(-x^2+3x+6\right)^2\)
\(\Leftrightarrow x^2-3x=x^4+9x^2+36+2\left(-3x^3+18x-6x^2\right)\)
\(\Leftrightarrow x^2-3x=x^4+9x^2+36-6x^3+36x-12x^2\)
\(\Leftrightarrow x^4+9x^2+36-6x^3+36x-12x^2-x^2+3x=0\)
\(\Leftrightarrow x^4-6x^3-4x^2+39x+36=0\)
\(\Leftrightarrow x^4-4x^3-2x^3+8x^2-12x^2+48x-9x+36=0\)
\(\Leftrightarrow\left(x^4-4x^3\right)-\left(2x^3-8x^2\right)-\left(12x^2-48x\right)-\left(9x-36\right)=0\)
\(\Leftrightarrow x^3\left(x-4\right)-2x^2\left(x-4\right)-12x\left(x-4\right)-9\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-2x^2-12x-9\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3+x^2-3x^2-3x-9x-9\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^2-3x-9\right)\left(x+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[\left(x-\frac{3}{2}\right)^2-\left(\sqrt{\frac{45}{4}}\right)^2\right]\left(x+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}+\frac{\sqrt{45}}{2}\right)\left(x-\frac{3}{2}-\frac{\sqrt{45}}{2}\right)\left(x+1\right)\left(x-4\right)=0\)
... bạn tự giải tiếp nha
Có thể làm theo cách sau :
\(\sqrt{x^2-3x}=-x^2+3x+6\)
\(\Leftrightarrow x^2-3x+\sqrt{x^2-3x}-6=0\)
Đặt \(\sqrt{x^2-3x}=a\) ( >= 0 ) . Ta có :
\(a^2+a-6=0\) . Rồi bn giải tiếp nha
