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 \(x \ne 3\)
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\(\eqalign{ & \Leftrightarrow {{\left( {3x - 1} \right)\left( {x + 3} \right)} \over {\left( {x - 1} \right)\left( {x + 3} \right)}} - {{\left( {2x + 5} \right)\left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( {x + 3} \right)}} = {{\left( {x - 1} \right)\left( {x + 3} \right)} \over {\left( {x - 1} \right)\left( {x + 3} \right)}} - {4 \over {\left( {x - 1} \right)\left( {x + 3} \right)}} \cr & \Leftrightarrow \left( {3x - 1} \right)\left( {x + 3} \right) - \left( {2x + 5} \right)\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right) - 4 \cr & \Leftrightarrow 3{x^2} + 9x - x - 3 - 2{x^2} + 2x - 5x + 5 = {x^2} + 3x - x - 3 - 4 \cr & \Leftrightarrow 3{x^2} - 2{x^2} - {x^2} + 9x - x + 2x - 5x - 3x + x = - 3 - 4 + 3 - 5 \cr & \Leftrightarrow 3x = - 9 \cr} \)
\( \Leftrightarrow x = - 3\) (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 \(x = - {3 \over {26}}\)