Question

[Yeji]

Giải các phương trình sau :

a) \(x^4-\left(x^2+2\right)=4\)

b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)

c) \(\frac{2x-10}{4}=5+\frac{2-3x}{6}\)

d) \(\frac{2x}{\left(x-3\right)\left(x+1\right)}+\frac{x}{2\left(x-3\right)}=\frac{x}{2x+2}\)

e) \(\left(\frac{x+2}{x}\right)^2+\left(\frac{x}{x+2}\right)^2=2\)

f) \(\left(x-a\right)\left(x+a\right)+2x+a^2=-1\)

g) \(\frac{x-a}{2a}+\frac{x-2a}{3a}+\frac{x-3a}{4a}+\frac{x-4a}{5a}=-4\)

h) \(\left(x^2-3x+4\right)^2=\left(x^2-2x+3\right)\left(x^2-4x+5\right)\)

i ) \(\frac{x^2-4x+12}{x^2-4x+6}=x^2-4x+8\)

Answer 1

[Do Thi Diem Quynhyka]

a) đặt \(t=x^2\)  (t\(\ge0\))

=>\(t^2-t-2=0\)=>\(\orbr{\begin{cases}t=2\\t=-1\left(loại\right)\end{cases}}\)

=>\(x^2=2\)=>\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

Answer 2

[Do Thi Diem Quynhyka]

a) \(\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)

b)\(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

c)\(x=\frac{47}{6}\)