Đề: giải phương trình \(\dfrac{x+1}{x+2}+\dfrac{3}{x-3}=\dfrac{2}{x^2-x-6}+1\)
Giải
ĐK: x \(\ne\) -2; x \(\ne\) 3.
MTC: x2 - x - 6 = (x + 2).(x - 3)
Quy đồng và khử mẫu:
(x + 1). (x - 3) + 3(x + 2) = 2 + (x + 2).(x - 3)
x2 + x + 3 = x2 + 5x + 8
4x = -5
<=> x = \(\dfrac{-5}{4}\) thỏa mãn ĐK.
Đặt :t=căn(x^2-2x+3),t>=căn2.Khi đó phương trình trở thnh : (x+1)t=x^2+1 <= >x^2+1-(x+1)t=0
Bây giờ ta thêm bớt , để được phương trình bậc 2 theo t có denta chẵn :
X^2-2x+3-(x+1)t+2(x-1)=0< =>t^2-(x+1)t+2(x-1)=0 <+ >t=2 hay t=x-1
Từ một phương trình đơn giản : (căn(1-x) -2.căn(1+x)). (căn(1-x) -2 –căn(1+x))=0, khai triển ra ta sẽ được pt sau
\(\dfrac{x+1}{x+2}+\dfrac{3}{x-3}=\dfrac{2}{x^2-x-6}+1\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\x-3\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)
\(\dfrac{x+1}{x+2}+\dfrac{3}{x-3}=\dfrac{2}{x^2-x-6}+1\\ \dfrac{x+1}{x+2}+\dfrac{3}{x-3}=\dfrac{2}{\left(x+2\right)\left(x-3\right)}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}+\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}+\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{2}{\left(x+2\right)\left(x-3\right)}\)
\(\Rightarrow x^2-2x-3+3x+6+x^2-x-6=2\\ \Leftrightarrow2x^2-5=0\\ \Leftrightarrow2x^2=5\\ \Leftrightarrow x^2=\dfrac{5}{2}\Rightarrow x=\pm\sqrt{\dfrac{5}{2}}\left(thõa\:mãn\right)\)
vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{\dfrac{5}{2}}\right\}\)
\(\dfrac{x+1}{x+2}+\dfrac{3}{x-3}=\dfrac{2}{x^2-x-6}+1\)
\(\Rightarrow\dfrac{x+1}{x+2}+\dfrac{3}{x-3}=\dfrac{2}{x^2-x-6}+1\left(đk:x\ne-2;x\ne3\right)\)
\(\Leftrightarrow\dfrac{x+1}{x+2}+\dfrac{3}{x-3}-\dfrac{2}{x^2-x-6}=1\)
\(\Leftrightarrow\dfrac{x+1}{x+2}+\dfrac{3}{x-3}-\dfrac{2}{x^2+2x-3x-6}=1\)
\(\Leftrightarrow\dfrac{x+1}{x+2}+\dfrac{3}{x-3}-\dfrac{2}{x\cdot\left(x+2\right)-3\left(x+2\right)}=1\)
\(\Leftrightarrow\dfrac{x+1}{x+2}+\dfrac{3}{x-3}-\dfrac{2}{\left(x-3\right)\cdot\left(x+2\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\cdot\left(x+1\right)+3\left(x+2\right)-2}{\left(x-3\right)\cdot\left(x+2\right)}=1\)
\(\Leftrightarrow\dfrac{x^2+x-3x-3+3x+6-2}{\left(x-3\right)\cdot\left(x+2\right)}=1\)
\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-3\right)\cdot\left(x+2\right)}=1\)
\(\Leftrightarrow x^2+x+1=\left(x-3\right)\cdot\left(x+2\right)\)
\(\Leftrightarrow x^2+x+1=x^2+2x-3x-6\)
\(\Leftrightarrow x+1=2x-3x-6\)
\(\Leftrightarrow x+1=-x-6\)
\(\Leftrightarrow x+x=-6-1\)
\(\Leftrightarrow2x=-7\)
\(\Rightarrow x=-\dfrac{7}{2}\)
