Question

[ღღ_Sunny_ღღ😘😘]

y² + \(\frac{4y^2}{\left(y-2\right)^2}\)= 5

#Nếu sai đề sửa hộ :3333

Answer 1

[Lê Tài Bảo Châu]

\(ĐKXĐ:y\ne2\)

\(y^2+\frac{4y^2}{\left(y-2\right)^2}=5\)

\(\Leftrightarrow\frac{y^2\left(y-2\right)^2+4y^2}{y^2-4y+4}=5\)

\(\Leftrightarrow\frac{y^4-4y^3+8y^2}{y^2-4y+4}=5\)

\(\Leftrightarrow y^4-4y^3+8y^2=5y^2-20y+20\)

\(\Leftrightarrow y^4-4y^3+3y^2+20y-20=0\)

\(\Leftrightarrow\left(y^4-3y^3+20y\right)-\left(y^3-3y^2+20\right)=0\)

\(\Leftrightarrow y\left(y^3-3y^2+20\right)-\left(y^3-3y^2+20\right)=0\)

\(\Leftrightarrow\left(y^3-3y^2+20\right)\left(y-1\right)=0\)

\(\Leftrightarrow\left[\left(y^3-5y^2+10y\right)+\left(2y^2-10y+20\right)\right]\left(y-1\right)=0\)

\(\Leftrightarrow\left[y\left(y^2-5y+10\right)+2\left(y^2-5y+10\right)\right]\left(y-1\right)=0\)

\(\Leftrightarrow\left(y+2\right)\left(y-1\right)\left(y^2-5y+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=-2\left(tm\right)\\y=1\left(tm\right)\end{cases}}\)hoặc \(y^2-5y+10=0\left(1\right)\)

Xét pt (1) \(\Leftrightarrow y^2-2.y.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+10=0\)

\(\Leftrightarrow\left(y-\frac{5}{2}\right)^2+\frac{15}{4}=0\)

Vì \(\left(y-\frac{5}{2}\right)^2\ge0;\forall y\)

\(\Rightarrow\left(y-\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

\(\Rightarrow pt\left(1\right)\)vô nghiệm

Vậy tập hợp nghiệm của pt \(S=\left\{1;-2\right\}\)