Giải phương trình:
\(\dfrac{x}{x^2-x-2}+\dfrac{3x}{x^2+3x-2}=1\)
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Những câu hỏi liên quan
8 tháng 2 2021
1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)
Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)
2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)
Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
Suy ra: \(9x^2+6x+24x+16=9x^2\)
\(\Leftrightarrow30x+16=0\)
\(\Leftrightarrow30x=-16\)
hay \(x=-\dfrac{8}{15}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)
TT
20 tháng 4 2021
PT 2
\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))
\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Rightarrow2x^2-3x+6=0\)
=> PT vô nghiệm.
N
20 tháng 11 2021
\(ĐK:x\ne3;x\ne2\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(\dfrac{x+1}{x-2}+1+\dfrac{x^2}{x-2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\\dfrac{x^2+x+2}{x-2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x^2+x+2=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
QD
1
XO
9 tháng 2 2023
ĐKXĐ : \(x\inℝ\)
Ta có : \(\dfrac{x^2+4x+5}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)
\(\Leftrightarrow1+\dfrac{5x}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)
\(\Leftrightarrow x.\left(\dfrac{5}{x^2-x+5}-\dfrac{3}{x^2-3x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{5}{x^2-x+5}=\dfrac{3}{x^2-3x+5}\left(1\right)\end{matrix}\right.\)
Phương trình (1) <=> 5(x2 - 3x + 5) = 3(x2 - x + 5)
<=> 2x2 - 12x + 10 = 0
<=> x2 - 6x + 5 = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Tập nghiệm \(S=\left\{0;1;5\right\}\)
30 tháng 12 2023
a: \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)
=>\(9x^2+12x+4-\left(9x^2-12x+4\right)-5x-38=0\)
=>\(9x^2+7x-34-9x^2+12x-4=0\)
=>19x-38=0
=>19x=38
=>x=38/19=2
b: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
=>\(x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
=>\(x^3+3x^2+12x-9=x^3+3x^2+3x+1\)
=>12x-9=3x+1
=>12x-3x=1+9
=>9x=10
=>x=10/9
5N
1
24 tháng 1
Sửa đề: \(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)
ĐKXĐ: \(x\notin\left\{0;-2\right\}\)
\(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)
=>\(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x\left(x+2\right)}=\dfrac{x+1}{x}\)
=>\(x\left(2x-1\right)+3x+2=\left(x+1\right)\left(x+2\right)\)
=>\(2x^2-x+3x+2=x^2+3x+2\)
=>\(2x^2+2x-x^2-3x=0\)
=>\(x^2-x=0\)
=>x(x-1)=0
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
5 tháng 2 2022
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
28 tháng 8 2021
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
ĐKXĐ: \(x\notin\left\{2;-1;\dfrac{-3\pm\sqrt{17}}{2}\right\}\)
\(\dfrac{x}{x^2-x-2}+\dfrac{3x}{x^2+3x-2}=1\)
=>\(\dfrac{x\left(x^2+3x-2\right)+3x\left(x^2-x-2\right)}{\left(x^2-x-2\right)\left(x^2+3x-2\right)}=1\)
=>\(\dfrac{x^3+3x^2-2x+3x^3-3x^2-6x}{\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2}=1\)
=>\(4x^3-8x=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)
=>\(4x\left(x^2-2\right)=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)
=>\(\left(x^2-2\right)^2-2x\left(x^2-2\right)-3x^2=0\)
=>\(\left(x^2-2\right)^2-3x\left(x^2-2\right)+x\left(x^2-2\right)-3x^2=0\)
=>\(\left(x^2-2\right)\left(x^2-2-3x\right)+x\left(x^2-2-3x\right)=0\)
=>\(\left(x^2+x-2\right)\left(x^2-3x-2\right)=0\)
=>\(\left(x+2\right)\left(x-1\right)\left(x^2-3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=1\left(nhận\right)\\x=\dfrac{3\pm\sqrt{17}}{2}\left(nhận\right)\end{matrix}\right.\)