Question

[phạm thị thịnh]

\(\dfrac{y+1}{y^{2^{ }}-5y}-\dfrac{y-5}{2y^{2^{ }}+10y}=\dfrac{y+25}{2y^{2^{ }}-50}\)

giúp mk giải phương trình bài trên chiều mai mk cần rùi

Answer 1

[Akai Haruma]

Lời giải:

ĐKXĐ: \(y\neq \pm 5; y\neq 0\)

Ta có:

\(\frac{y+1}{y^2-5y}-\frac{y-5}{2y^2+10y}=\frac{y+25}{2y^2-50}\)

\(\Leftrightarrow \frac{y+1}{y(y-5)}-\frac{y-5}{2y(y+5)}-\frac{y+25}{2(y^2-25)}=0\)

\(\Leftrightarrow \frac{2(y+1)(y+5)}{2y(y-5)(y+5)}-\frac{(y-5)(y-5)}{2y(y+5)(y-5)}-\frac{y(y+25)}{2y(y^2-25)}=0\)

\(\Leftrightarrow \frac{2(y^2+6y+5)}{2y(y^2-25)}-\frac{y^2-10y+25}{2y(y^2-25)}-\frac{y^2+25y}{2y(y^2-25)}=0\)

\(\Leftrightarrow \frac{2(y^2+6y+5)-(y^2-10y+25)-(y^2+25y)}{2y(y^2-25)}=0\)

\(\Leftrightarrow \frac{-3(y+5)}{2y(y^2-25)}=0\)

\(\Leftrightarrow -3(y+5)=0\Leftrightarrow y+5=0\Leftrightarrow y=-5\) (không t/m ĐKXĐ)

Vậy PT vô nghiệm.

Answer 2

[Gia Hân Ngô]

\(\dfrac{y+1}{y^2-5y}-\dfrac{y-5}{2y^2+10y}=\dfrac{y+25}{2y^2-50}\left(ĐKXĐ:y\ne O;y\ne\pm5\right)\)

\(\Leftrightarrow\dfrac{y+1}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)

\(\Leftrightarrow\dfrac{2\left(y+1\right)\left(y+5\right)-\left(y-5\right)^2}{2y\left(y-5\right)\left(y+5\right)}=\dfrac{y\left(y+25\right)}{2y\left(y-5\right)\left(y+5\right)}\)

\(\Rightarrow2\left(y+1\right)\left(y+5\right)-\left(y-5\right)^2=y\left(y+25\right)\)

\(\Leftrightarrow\left(2y+2\right)\left(y+5\right)-\left(y^2-10y+25\right)=y^2+25y\)

\(\Leftrightarrow2y^2+10y+2y+10-y^2+10y-25=y^2+25y\)

\(\Leftrightarrow y^2+22y-15=y^2+25y\)

\(\Leftrightarrow y^2-y^2+22y-25y=15\)

\(\Leftrightarrow-3y=15\)

\(\Leftrightarrow y=-5\) (ko thỏa mãn ĐKXĐ)

Vậy ....................