Giải các phương trình sau:
a. \({{1 – x} \over {x + 1}} + 3 = {{2x + 3} \over {x + 1}}\)
b. \({{{{\left( {x + 2} \right)}^2}} \over {2x – 3}} – 1 = {{{x^2} + 10} \over {2x – 3}}\)
c. \({{5x – 2} \over {2 – 2x}} + {{2x – 1} \over 2} = 1 – {{{x^2} + x – 3} \over {1 – x}}\)
d. \({{5 – 2x} \over 3} + {{\left( {x – 1} \right)\left( {x + 1} \right)} \over {3x – 1}} = {{\left( {x + 2} \right)\left( {1 – 3x} \right)} \over {9x – 3}}\)
a. \({{1 – x} \over {x + 1}} + 3 = {{2x + 3} \over {x + 1}}\) ĐKXĐ: \(x \ne – 1\)
\(\eqalign{ & \Leftrightarrow {{1 – x} \over {x + 1}} + {{3\left( {x + 1} \right)} \over {x + 1}} = {{2x + 3} \over {x + 1}} \cr & \Leftrightarrow 1 – x + 3\left( {x + 1} \right) = 2x + 3 \cr & \Leftrightarrow 1 – x + 3x + 3 – 2x – 3 = 0 \cr & \Leftrightarrow 0x = – 1 \cr} \)
Phương trình vô nghiệm.
b. \({{{{\left( {x + 2} \right)}^2}} \over {2x – 3}} – 1 = {{{x^2} + 10} \over {2x – 3}}\)
ĐKXĐ: \(x \ne {3 \over 2}\)
\(\eqalign{ & \Leftrightarrow {{{{\left( {x + 2} \right)}^2}} \over {2x – 3}} – {{2x – 3} \over {2x – 3}} = {{{x^2} + 10} \over {2x – 3}} \cr & \Leftrightarrow {\left( {x + 2} \right)^2} – \left( {2x – 3} \right) = {x^2} + 10 \cr & \Leftrightarrow {x^2} + 4x + 4 – 2x + 3 – {x^2} – 10 = 0 \cr & \Leftrightarrow 2x = 3 \cr} \)
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\( \Leftrightarrow x = {3 \over 2}\) (loại)
Phương trình vô nghiệm.
c. \({{5x – 2} \over {2 – 2x}} + {{2x – 1} \over 2} = 1 – {{{x^2} + x – 3} \over {1 – x}}\)
ĐKXĐ: \(x \ne 1\)
\(\eqalign{ & \Leftrightarrow {{5x – 2} \over {2\left( {1 – x} \right)}} + {{\left( {2x – 1} \right)\left( {1 – x} \right)} \over {2\left( {1 – x} \right)}} = {{2\left( {1 – x} \right)} \over {2\left( {1 – x} \right)}} – {{2\left( {{x^2} + x – 3} \right)} \over {2\left( {1 – x} \right)}} \cr & \Leftrightarrow 5x – 2 + \left( {2x – 1} \right)\left( {1 – x} \right) = 2\left( {1 – x} \right) – 2\left( {{x^2} + x – 3} \right) \cr & \Leftrightarrow 5x – 2 + 2x – 2{x^2} – 1 + x – 2 + 2x + 2{x^2} + 2x – 6 = 0 \cr & \Leftrightarrow 5x + 2x + x + 2x + 2x = 2 + 6 + 2 + 1 \Leftrightarrow 12x = 11 \cr} \)
\( \Leftrightarrow x = {{11} \over {12}}\) (thỏa)
Vậy phương trình có nghiệm \(x = {{11} \over {12}}\)
d. \({{5 – 2x} \over 3} + {{\left( {x – 1} \right)\left( {x + 1} \right)} \over {3x – 1}} = {{\left( {x + 2} \right)\left( {1 – 3x} \right)} \over {9x – 3}}\) ĐKXĐ: \(x \ne {1 \over 3}\)
\(\eqalign{ & \Leftrightarrow {{\left( {5 – 2x} \right)\left( {3x – 1} \right)} \over {3\left( {3x – 1} \right)}} + {{3\left( {x + 1} \right)\left( {x – 1} \right)} \over {3\left( {3x – 1} \right)}} = {{\left( {x + 2} \right)\left( {1 – 3x} \right)} \over {3\left( {3x – 1} \right)}} \cr & \Leftrightarrow \left( {5 – 2x} \right)\left( {3x – 1} \right) + 3\left( {x + 1} \right)\left( {x – 1} \right) = \left( {x + 2} \right)\left( {1 – 3x} \right) \cr & \Leftrightarrow 15x – 5 – 6{x^2} + 2x + 3{x^2} – 3 = x – 3{x^2} + 2 – 6x \cr & \Leftrightarrow – 6{x^2} + 3{x^2} + 3{x^2} + 15x + 2x – x + 6x = 2 + 5 + 3 \cr & \Leftrightarrow 22x = 10 \cr} \)
\( \Leftrightarrow x = {5 \over {11}}\) (thỏa)
Vậy phương trình có nghiệm \(x = {5 \over {11}}\)