Question

[Trần Trí Hùng]

giúp mình nhé

Answer 1

[Yen Nhi]

Answer:

Bài 1:

a) \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{x^2+2x-3}=1\)

\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{4}{\left(x-1\right).\left(x+3\right)}=1\left(ĐK:x\ne1;x\ne-3\right)\)

\(\left(3x-1\right).\left(x+3\right)-\left(2x+5\right).\left(x-1\right)+4=\left(x+3\right).\left(x-1\right)\)

\(3x^2+8x-3-2x^2-5x+2x+5+4=x^2+2x-3\)

\(3x+9=0\Rightarrow x=-3\) (Không thoả mãn)

Vậy phương trình vô nghiệm

b) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(ĐK:x\ne1\right)\)

\(\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}-\frac{4\left(x-1\right)}{x^3-1}=0\)

\(\frac{x^2+x+1+2x^2-5-4x+4}{x^3-1}=0\)

\(3x^2-3x=0\)

\(3x.\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\text{(Thoả mãn)}\\x=1\text{(Loại)}\end{cases}}\)