Question

[Minh Đỗ]

Giải các phương trình:

a)\(\left(x^2-5x+1\right)\left(x^2-4\right)=6\left(x-1\right)^2\)

b)\(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

Answer 1

[Thắng Nguyễn]

(x²-5x+1)(x²-4)=6(x-1)²

<=>(x²-5x+1)(x²-4)-6(x-1)²=0

<=>x⁴-5x³-3x²+20x-4-6x²+12x-6=0

<=>x⁴-5x³-9x²+32x-10=0

<=>x⁴-6x³+2x²+x³-6x²+2x-5x²+30x-10=0

<=>x²(x²-6x+2)+x(x²-6x+2)-5(x²-6x+2)=0

<=>(x²-6x+2)(x²+x-5)=0

• Với x²-6x+2=0 <=>x²-6x+9-7=0

<=>(x-3)²-7=0

\(\Leftrightarrow x-3=-\sqrt{7}hoac\sqrt{7}\)

\(\Leftrightarrow3\pm\sqrt{7}\)

• Với x²+x-5=0 <=>\(\left(x+\frac{1}{2}\right)^2-\frac{21}{4}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{21}{4}\)

\(\Leftrightarrow x=-\frac{1}{2}-\frac{\sqrt{21}}{2}hoac\frac{\sqrt{21}}{2}-\frac{1}{2}\)

b)\(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)

tính mẫu ra rồi rút gọn,x=1