Ta có:\(\dfrac{x^2-10+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}\left(đkxđ:x\ne\sqrt{21}+6;-\sqrt{21}+6\right)\)
\(\Leftrightarrow\dfrac{x^2-6x+15-4x}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}\)
\(\Leftrightarrow1-\dfrac{4x}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}\)
\(\Leftrightarrow\dfrac{4x}{x^2-6x+15}+\dfrac{4x}{x^2-12x+15}=1\)
\(\Leftrightarrow\dfrac{4}{x-6+\dfrac{15}{x}}+\dfrac{4}{x-12+\dfrac{15}{x}}=1\)
Đặt \(x+\dfrac{15}{x}=t\)
PT\(\Leftrightarrow\dfrac{4}{t-6}+\dfrac{4}{t-12}=1\)
\(\Leftrightarrow4t-48+4t-24=t^2-18t+72\)
\(\Leftrightarrow8t-72=t^2-18t+72\)
\(\Leftrightarrow t^2-26t+144=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=18\\t=8\end{matrix}\right.\)
Thay vào từng trường hợp rồi tìm x
\(\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{4x}{x^2-12x+15}\)
đặt :\(x^2-6x+15=y\) ta đc:
\(\dfrac{y^2-4x}{y}=\dfrac{4x}{y^2-6x}\)
<=>\(\dfrac{\left(y^2-4x\right)\left(y^2-6x\right)}{y\left(y^2-6x\right)}=\dfrac{4xy}{y\left(y^2-6x\right)}\)
=>\(y^4-6xy^2-4xy^2+24x^2=4xy\)
<=>
tớ xin lỗi, tớ k làm đc định hủy thì ấn nhầm vào trả lời, xin lỗi nhiều
