a)Cho hai biểu thức A=(5/2x+1) và B=(4/2x-1). Hãy tìm các giá trị của x để hai biểu thức ấy có giá trị thỏa mãn hệ thức:
a) 2A+3B=0 b)AB=A+B
b)Cho phương trình (x+m/x-m)-(x-m/x+m)=(m(3m+1)/m^2-x^2)(m là tham số). Giải phương trình khi:
a)m=-3 b)m=0
Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ne – \dfrac{1}{2}\\
x \ne \dfrac{1}{2}
\end{array} \right.\\
1)2.A + 3B = 0\\
\Rightarrow 2.\dfrac{5}{{2x + 1}} + 3.\dfrac{4}{{2x – 1}} = 0\\
\Rightarrow \dfrac{{10\left( {2x – 1} \right) + 12\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {2x – 1} \right)}} = 0\\
\Rightarrow 5.\left( {2x – 1} \right) + 6\left( {2x + 1} \right) = 0\\
\Rightarrow 10x – 5 + 12x + 6 = 0\\
\Rightarrow 22x = – 1\\
\Rightarrow x = – \dfrac{1}{{22}}\left( {tmdk} \right)\\
Vậy\,x = – \dfrac{1}{{22}}\\
b)AB = A + B\\
\Rightarrow \dfrac{5}{{2x + 1}}.\dfrac{4}{{2x – 1}} = \dfrac{5}{{2x + 1}} + \dfrac{4}{{2x – 1}}\\
\Rightarrow \dfrac{{20}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}} = \dfrac{{5\left( {2x – 1} \right) + 4\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {2x – 1} \right)}}\\
\Rightarrow 20 = 10x – 5 + 8x + 4\\
\Rightarrow 18x = 21\\
\Rightarrow x = \dfrac{7}{6}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{7}{6}\\
b)\dfrac{{x + m}}{{x – m}} – \dfrac{{x – m}}{{x + m}} = \dfrac{{m\left( {3m + 1} \right)}}{{{m^2} – {x^2}}}\\
\Rightarrow \dfrac{{{{\left( {x + m} \right)}^2} – {{\left( {x – m} \right)}^2}}}{{\left( {x + m} \right)\left( {x – m} \right)}} + \dfrac{{3{m^2} + m}}{{{x^2} – {m^2}}} = 0\\
\Rightarrow \dfrac{{4mx + 3{m^2} + m}}{{\left( {x + m} \right)\left( {x – m} \right)}} = 0\\
\Rightarrow 4mx + 3{m^2} + m = 0\\
a)m = – 3\\
4.\left( { – 3} \right).x + 3.{\left( { – 3} \right)^2} + \left( { – 3} \right) = 0\\
\Rightarrow – 12x + 24 = 0\\
\Rightarrow x = 2\left( {tmdk} \right)\\
b)m = 0\\
\Rightarrow 4.0.x + 0 + 0 = 0
\end{array}$
Vậy pt nghiệm đúng với mọi x#0 khi m=0
$\begin{array}{l} a)Dkxd:\left\{ \begin{array}{l} x \ne – \dfrac{1}{2}\\ x \ne \dfrac{1}{2} \end{array} \right.\\ 1)2.A + 3B = 0\\ \Rightarrow 2.\dfrac{5}{{2x + 1}} + 3.\dfrac{4}{{2x – 1}} = 0\\ \Rightarrow \dfrac{{10\left( {2x – 1} \right) + 12\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {2x – 1} \right)}} = 0\\ \Rightarrow 5.\left( {2x – 1} \right) + 6\left( {2x + 1} \right) = 0\\ \Rightarrow 10x – 5 + 12x + 6 = 0\\ \Rightarrow 22x = – 1\\ \Rightarrow x = – \dfrac{1}{{22}}\left( {tmdk} \right)\\ Vậy\,x = – \dfrac{1}{{22}}\\ b)AB = A + B\\ \Rightarrow \dfrac{5}{{2x + 1}}.\dfrac{4}{{2x – 1}} = \dfrac{5}{{2x + 1}} + \dfrac{4}{{2x – 1}}\\ \Rightarrow \dfrac{{20}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}} = \dfrac{{5\left( {2x – 1} \right) + 4\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {2x – 1} \right)}}\\ \Rightarrow 20 = 10x – 5 + 8x + 4\\ \Rightarrow 18x = 21\\ \Rightarrow x = \dfrac{7}{6}\left( {tmdk} \right)\\ Vậy\,x = \dfrac{7}{6}\\ b)\dfrac{{x + m}}{{x – m}} – \dfrac{{x – m}}{{x + m}} = \dfrac{{m\left( {3m + 1} \right)}}{{{m^2} – {x^2}}}\\ \Rightarrow \dfrac{{{{\left( {x + m} \right)}^2} – {{\left( {x – m} \right)}^2}}}{{\left( {x + m} \right)\left( {x – m} \right)}} + \dfrac{{3{m^2} + m}}{{{x^2} – {m^2}}} = 0\\ \Rightarrow \dfrac{{4mx + 3{m^2} + m}}{{\left( {x + m} \right)\left( {x – m} \right)}} = 0\\ \Rightarrow 4mx + 3{m^2} + m = 0\\ a)m = – 3\\ 4.\left( { – 3} \right).x + 3.{\left( { – 3} \right)^2} + \left( { – 3} \right) = 0\\ \Rightarrow – 12x + 24 = 0\\ \Rightarrow x = 2\left( {tmdk} \right)\\ b)m = 0\\ \Rightarrow 4.0.x + 0 + 0 = 0 \end{array}$