Bài 1: Giải phương trình sau:
a) \(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
b) \(\frac{2x+3}{2x-3}-\frac{3}{4x-6}=\frac{2}{5}\)
c) \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)
d) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{7x-3}{9-x^2}\)
e) \(\frac{1}{x+1}+\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}\)
Bài 2: Giải phương trình:
(x2 - 3)2 + 2(x2 - 3) - 3 = 0
Giúp tớ với nhé các cậu :)))
Bài 1:
a/ \(x\ne1;2\)
\(\frac{x-2}{\left(x-1\right)\left(x-2\right)}-\frac{7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x-2-7x+7+1=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Rightarrow x=1\) (loại)
Vậy pt vô nghiệm
b/ \(x\ne\frac{3}{2}\)
\(\frac{2x+3}{2x-3}-\frac{3}{2\left(2x-3\right)}-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{10\left(2x+3\right)}{10\left(2x-3\right)}-\frac{15}{10\left(2x-3\right)}-\frac{4\left(2x-3\right)}{10\left(2x-3\right)}=0\)
\(\Leftrightarrow20x+30-15-8x+12=0\)
\(\Leftrightarrow12x+27=0\)
\(\Rightarrow x=-\frac{9}{4}\)
c/ \(x\ne\pm1\)
\(\frac{x+1}{x-1}-\frac{4}{x+1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow x^2+2x+1-4x+4+3-x^2=0\)
\(\Leftrightarrow-2x+8=0\)
\(\Rightarrow x=4\)
Bài 1:
d/\(x\ne\pm3\)
\(\frac{x-1}{x+3}-\frac{x}{x-3}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)
\(\Rightarrow0=0\)
Vậy pt có vô số nghiệm \(x\ne\pm3\)
e/ \(x\ne\pm1\)
\(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Leftrightarrow x^2-2x+1+2+3x-3=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\left(l\right)\end{matrix}\right.\)
Bài 2:
Đặt \(x^2-3=a\) phương trình trở thành:
\(a^2+2a-3=0\)
\(\Leftrightarrow a^2-a+3a-3=0\)
\(\Leftrightarrow a\left(a-1\right)+3\left(a-1\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(a-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=0\end{matrix}\right.\)
