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Giải các phương trình sau:
a. \({{9x – 0,7} \over 4} – {{5x – 1,5} \over 7} = {{7x – 1,1} \over 3} – {{5\left( {0,4 – 2x} \right)} \over 6}\)
b. \({{3x – 1} \over {x – 1}} – {{2x + 5} \over {x + 3}} = 1 – {4 \over {\left( {x – 1} \right)\left( {x + 3} \right)}}\)
c. \({3 \over {4\left( {x – 5} \right)}} + {{15} \over {50 – 2{x^2}}} = – {7 \over {6\left( {x + 5} \right)}}\)
d. \({{8{x^2}} \over {3\left( {1 – 4{x^2}} \right)}} = {{2x} \over {6x – 3}} – {{1 + 8x} \over {4 + 8x}}\)
a. \({{9x – 0,7} \over 4} – {{5x – 1,5} \over 7} = {{7x – 1,1} \over 3} – {{5\left( {0,4 – 2x} \right)} \over 6}\)
\( \Leftrightarrow {{21\left( {9x – 0,7} \right)} \over {84}} – {{12\left( {5x – 1,5} \right)} \over {84}}\) = \({{28\left( {7x – 1,1} \right)} \over {84}} – {{70\left( {0,4 – 2x} \right)} \over {84}}\)
\(\eqalign{ & \Leftrightarrow 21\left( {9x – 0,7} \right) – 12\left( {5x – 1,5} \right) = 28\left( {7x – 1,1} \right) – 70\left( {0,4 – 2x} \right) \cr & \Leftrightarrow 189x – 14,7 – 60x + 18 = 196x – 30,8 – 28 + 140x \cr & \Leftrightarrow 189x – 60x – 196x – 140x = – 30,8 – 28 + 14,7 – 18 \cr & \Leftrightarrow – 207x = – 62,1 \cr & \Leftrightarrow x = 0,3 \cr} \)
Vậy phương trình có nghiệm x = 0,3
b. \({{3x – 1} \over {x – 1}} – {{2x + 5} \over {x + 3}} = 1 – {4 \over {\left( {x – 1} \right)\left( {x + 3} \right)}}\) ĐKXĐ: \(x \ne 1\)và \(x \ne 3\)
\(\eqalign{ & \Leftrightarrow {{\left( {3x – 1} \right)\left( {x + 3} \right)} \over {\left( {x – 1} \right)\left( {x + 3} \right)}} – {{\left( {2x + 5} \right)\left( {x – 1} \right)} \over {\left( {x – 1} \right)\left( {x + 3} \right)}} = {{\left( {x – 1} \right)\left( {x + 3} \right)} \over {\left( {x – 1} \right)\left( {x + 3} \right)}} – {4 \over {\left( {x – 1} \right)\left( {x + 3} \right)}} \cr & \Leftrightarrow \left( {3x – 1} \right)\left( {x + 3} \right) – \left( {2x + 5} \right)\left( {x – 1} \right) = \left( {x – 1} \right)\left( {x + 3} \right) – 4 \cr & \Leftrightarrow 3{x^2} + 9x – x – 3 – 2{x^2} + 2x – 5x + 5 = {x^2} + 3x – x – 3 – 4 \cr & \Leftrightarrow 3{x^2} – 2{x^2} – {x^2} + 9x – x + 2x – 5x – 3x + x = – 3 – 4 + 3 – 5 \cr & \Leftrightarrow 3x = – 9 \cr} \)
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\( \Leftrightarrow x = – 3\) (loại)
Vậy phương trình vô nghiệm
c. \({3 \over {4\left( {x – 5} \right)}} + {{15} \over {50 – 2{x^2}}} = – {7 \over {6\left( {x + 5} \right)}}\) ĐKXĐ: \(x \ne \pm 5\)
\(\eqalign{ & \Leftrightarrow {3 \over {4\left( {x – 5} \right)}} + {{15} \over {2\left( {25 – {x^2}} \right)}} = – {7 \over {6\left( {x + 5} \right)}} \cr & \Leftrightarrow {3 \over {4\left( {x – 5} \right)}} – {{15} \over {2\left( {x + 5} \right)\left( {x – 5} \right)}} = – {7 \over {6\left( {x + 5} \right)}} \cr & \Leftrightarrow {{9\left( {x + 5} \right)} \over {12\left( {x + 5} \right)\left( {x – 5} \right)}} – {{90} \over {12\left( {x + 5} \right)\left( {x – 5} \right)}} = – {{14\left( {x – 5} \right)} \over {12\left( {x + 5} \right)\left( {x – 5} \right)}} \cr & \Leftrightarrow 9\left( {x + 5} \right) – 90 = – 14\left( {x – 5} \right) \cr & \Leftrightarrow 9x + 45 – 90 = – 14x + 70 \cr & \Leftrightarrow 9x + 14x = 70 – 45 + 90 \cr & \Leftrightarrow 23x = 115 \cr} \)
\( \Leftrightarrow x = 5\) (loại)
Vậy phương trìnhvô nghiệm
d. \({{8{x^2}} \over {3\left( {1 – 4{x^2}} \right)}} = {{2x} \over {6x – 3}} – {{1 + 8x} \over {4 + 8x}}\) ĐKXĐ: \(x \ne \pm {1 \over 2}\)
\(\eqalign{ & \Leftrightarrow {{8{x^2}} \over {3\left( {1 – 2x} \right)\left( {1 + 2x} \right)}} = {{ – 2x} \over {3\left( {1 – 2x} \right)}} – {{1 + 8x} \over {4\left( {1 + 2x} \right)}} \cr & \Leftrightarrow {{32{x^2}} \over {12\left( {1 – 2x} \right)\left( {1 + 2x} \right)}} = {{ – 8x\left( {1 + 2x} \right)} \over {12\left( {1 – 2x} \right)\left( {1 + 2x} \right)}} – {{3\left( {1 + 8x} \right)\left( {1 – 2x} \right)} \over {12\left( {1 – 2x} \right)\left( {1 + 2x} \right)}} \cr & \Leftrightarrow 32{x^2} = – 8x – 16{x^2} – 3\left( {1 – 2x + 8x – 16{x^2}} \right) \cr & \Leftrightarrow 32{x^2} = – 8x – 16{x^2} – 3 – 18x + 48{x^2} \cr & \Leftrightarrow 32{x^2} + 16{x^2} – 48{x^2} + 18x + 8x = – 3 \cr & \Leftrightarrow 26x = – 3 \cr} \)
\( \Leftrightarrow x = – {3 \over {26}}\) (thỏa mãn)
Vậy phương trình có nghiệm \(x = – {3 \over {26}}\)