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Giải các phương trình sau:
a. \({{2x + 1} \over {x – 1}} = {{5\left( {x – 1} \right)} \over {x + 1}}\)
b. \({{x – 3} \over {x – 2}} + {{x – 2} \over {x – 4}} = – 1\)
c. \({1 \over {x – 1}} + {{2{x^2} – 5} \over {{x^3} – 1}} = {4 \over {{x^2} + x + 1}}\)
d. \({{13} \over {\left( {x – 3} \right)\left( {2x + 7} \right)}} + {1 \over {2x + 7}} = {6 \over {{x^2} – 9}}\)
a. \({{2x + 1} \over {x – 1}} = {{5\left( {x – 1} \right)} \over {x + 1}}$ ĐKXĐ:
\(\eqalign{ & \Leftrightarrow {{\left( {2x + 1} \right)\left( {x + 1} \right)} \over {\left( {x + 1} \right)\left( {x – 1} \right)}} = {{5\left( {x – 1} \right)\left( {x – 1} \right)} \over {\left( {x + 1} \right)\left( {x – 1} \right)}} \cr & \Leftrightarrow \left( {2x + 1} \right)\left( {x + 1} \right) = 5\left( {x – 1} \right)\left( {x – 1} \right) \cr & \Leftrightarrow 2{x^2} + 2x + x + 1 = 5{x^2} – 10x + 5 \cr & \Leftrightarrow 2{x^2} – 5{x^2} + 2x + x + 10x + 1 – 5 = 0 \cr & \Leftrightarrow – 3{x^2} + 13x – 4 = 0 \cr & \Leftrightarrow 3{x^2} – x – 12x + 4 = 0 \cr & \Leftrightarrow x\left( {3x – 1} \right) – 4\left( {3x – 1} \right) = 0 \cr & \Leftrightarrow \left( {3x – 1} \right)\left( {x – 4} \right) = 0 \cr} \)
\( \Leftrightarrow x – 4 = 0\) hoặc \(3x – 1 = 0\)
+) \(x – 4 = 0 \Leftrightarrow x = 4\) (thỏa mãn)
+) \(3x – 1 = 0 \Leftrightarrow x = {1 \over 3}\) (thỏa mãn)
Vậy phương trình có nghiệm x = 4 hoặc \(x = {1 \over 3}\)
b. \({{x – 3} \over {x – 2}} + {{x – 2} \over {x – 4}} = – 1\) ĐKXĐ: \(x \ne 2\)và \(x \ne 4\)
\(\eqalign{ & \Leftrightarrow {{\left( {x – 3} \right)\left( {x – 4} \right)} \over {\left( {x – 2} \right)\left( {x – 4} \right)}} + {{\left( {x – 2} \right)\left( {x – 2} \right)} \over {\left( {x – 2} \right)\left( {x – 4} \right)}} = – {{\left( {x – 2} \right)\left( {x – 4} \right)} \over {\left( {x – 2} \right)\left( {x – 4} \right)}} \cr & \Leftrightarrow \left( {x – 3} \right)\left( {x – 4} \right) + \left( {x – 2} \right)\left( {x – 2} \right) = – \left( {x – 2} \right)\left( {x – 4} \right) \cr & \Leftrightarrow {x^2} – 4x – 3x + 12 + {x^2} – 2x – 2x + 4 = – {x^2} + 4x + 2x – 8 \cr & \Leftrightarrow 3{x^2} – 17x + 24 = 0 \cr & \Leftrightarrow 3{x^2} – 9x – 8x + 24 = 0 \cr & \Leftrightarrow 3x\left( {x – 3} \right) – 8\left( {x – 3} \right) = 0 \cr & \Leftrightarrow \left( {3x – 8} \right)\left( {x – 3} \right) = 0 \cr} \)
\( \Leftrightarrow 3x – 8 = 0\) hoặc \(x – 3 = 0\)
+ \(3x – 8 = 0 \Leftrightarrow x = {8 \over 3}\) (thỏa mãn)
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+ \(x – 3 = 0 \Leftrightarrow x = 3\) (thỏa mãn)
Vậy phương trình có nghiệm \(x = {8 \over 3}\) hoặc x = 3
c. \({1 \over {x – 1}} + {{2{x^2} – 5} \over {{x^3} – 1}} = {4 \over {{x^2} + x + 1}}\)
ĐKXĐ: \(x \ne 1\)
\(\eqalign{ & \Leftrightarrow {{{x^2} + x + 1} \over {{x^3} – 1}} + {{2{x^2} – 5} \over {{x^3} – 1}} = {{4\left( {x – 1} \right)} \over {{x^3} – 1}} \cr & \Leftrightarrow {x^2} + x + 1 + 2{x^2} – 5 = 4\left( {x – 1} \right) \cr & \Leftrightarrow {x^2} + x + 1 + 2{x^2} – 5 = 4x – 4 \cr & \Leftrightarrow {x^2} + 2{x^2} + x – 4x = – 4 + 5 – 1 \cr & \Leftrightarrow 3{x^2} – 3x = 0 \cr & \Leftrightarrow 3x\left( {x – 1} \right) = 0 \cr} \)
\( \Leftrightarrow x = 0\) (thỏa) hoặc \(x – 1 = 0 \Leftrightarrow x = 1\) (loại)
Vậy phương trình có nghiệm x = 0
d. \({{13} \over {\left( {x – 3} \right)\left( {2x + 7} \right)}} + {1 \over {2x + 7}} = {6 \over {{x^2} – 9}}\) ĐKXĐ: \(x \ne \pm 3\) và \(x = – {7 \over 2}\)
\(\eqalign{ & \Leftrightarrow {{13\left( {x + 3} \right)} \over {\left( {{x^2} – 9} \right)\left( {2x + 7} \right)}} + {{{x^2} – 9} \over {\left( {{x^2} – 9} \right)\left( {2x + 7} \right)}} = {{6\left( {2x + 7} \right)} \over {\left( {{x^2} – 9} \right)\left( {2x + 7} \right)}} \cr & \Leftrightarrow 13\left( {x + 3} \right) + {x^2} – 9 = 6\left( {2x + 7} \right) \cr & \Leftrightarrow 13x + 39 + {x^2} – 9 = 12x + 42 \cr & \Leftrightarrow {x^2} + x – 12 = 0 \cr & \Leftrightarrow {x^2} – 3x + 4x – 12 = 0 \cr & \Leftrightarrow x\left( {x – 3} \right) + 4\left( {x – 3} \right) = 0 \cr & \Leftrightarrow \left( {x + 4} \right)\left( {x – 3} \right) = 0 \cr} \)
\( \Leftrightarrow x + 4 = 0\) hoặc \(x – 3 = 0\)
+ \(x + 4 = 0 \Leftrightarrow x = – 4\) (thỏa mãn)
+ \(x – 3 = 0 \Leftrightarrow x = 3\) (loại)
Vậy phương trình có nghiệm x = -4