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Bài 57. Giải các phương trình:
a) \(5{{\rm{x}}^2} – 3{\rm{x}} + 1 = 2{\rm{x}} + 11\)
b) \({{{x^2}} \over 5} – {{2{\rm{x}}} \over 3} = {{x + 5} \over 6}\)
c) \({x \over {x – 2}} = {{10 – 2{\rm{x}}} \over {{x^2} – 2{\rm{x}}}}\)
d) \({{x + 0,5} \over {3{\rm{x}} + 1}} = {{7{\rm{x}} + 2} \over {9{{\rm{x}}^2} – 1}}\)
e) \(2\sqrt 3 {x^2} + x + 1 = \sqrt 3 \left( {x + 1} \right)\)
f) \({x^2} + 2\sqrt 2 x + 4 = 3\left( {x + \sqrt 2 } \right)\)
Hướng dẫn làm bài:
a)
\(\eqalign{
& 5{{\rm{x}}^2} – 3{\rm{x}} + 1 = 2{\rm{x}} + 11 \cr
& \Leftrightarrow 5{{\rm{x}}^2} – 5{\rm{x}} – 10 = 0 \cr
& \Leftrightarrow {x^2} – x – 2 = 0 \cr}\)
Phương trình có \(a – b + c = 1 + 1 – 2 = 0\) nên có 2 nghiệm \({x_1}= -1; {x_2}= 2\)
b)
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\(\eqalign{
& {{{x^2}} \over 5} – {{2{\rm{x}}} \over 3} = {{x + 5} \over 6} \cr
& \Leftrightarrow 6{{\rm{x}}^2} – 20{\rm{x}} = 5{\rm{x}} + 25 \cr
& \Leftrightarrow 6{{\rm{x}}^2} – 25{\rm{x}} – 25 = 0 \cr
& \Delta = {25^2} + 4.6.25 = 1225 \cr
& \sqrt \Delta = 35 \Rightarrow {x_1} = 5;{x_2} = – {5 \over 6} \cr} \)
c) \({x \over {x – 2}} = {{10 – 2{\rm{x}}} \over {{x^2} – 2{\rm{x}}}}\) ĐKXĐ: \(x ≠ 0; x ≠ 2\)
\(\eqalign{
& \Leftrightarrow {x^2} = 10 – 2{\rm{x}} \cr
& \Leftrightarrow {x^2} + 2{\rm{x}} – 10 = 0 \cr
& \Delta ‘ = 1 + 10 = 11 \cr
& \Rightarrow {x_1} = – 1 + \sqrt {11} (TM) \cr
& {x_2} = – 1 – \sqrt {11} (TM) \cr} \)
d) \({{x + 0,5} \over {3{\rm{x}} + 1}} = {{7{\rm{x}} + 2} \over {9{{\rm{x}}^2} – 1}}\) ĐKXĐ: \(x \ne \pm {1 \over 3}\)
\(\eqalign{
& \Leftrightarrow {{2{\rm{x}} + 1} \over {3{\rm{x}} + 1}} = {{14{\rm{x}} + 4} \over {9{{\rm{x}}^2} – 1}} \cr
& \Leftrightarrow \left( {2{\rm{x}} + 1} \right)\left( {3{\rm{x}} – 1} \right) = 14{\rm{x}} + 4 \cr
& \Leftrightarrow 6{{\rm{x}}^2} + x – 1 = 14{\rm{x}} + 4 \cr
& \Leftrightarrow 6{{\rm{x}}^2} – 13{\rm{x}} – 5 = 0 \cr
& \Delta = {( – 13)^2} – 4.6.( – 5) = 289 \cr
& \sqrt \Delta = \sqrt {289} = 17 \cr
& \Rightarrow {x_1} = {5 \over 2}(TM) \cr
& {x_2} = – {1 \over 3}(loại) \cr} \)
e)
\(\eqalign{
& 2\sqrt 3 {x^2} + x + 1 = \sqrt 3 \left( {x + 1} \right) \cr
& \Leftrightarrow 2\sqrt 3 {x^2} – \left( {\sqrt 3 – 1} \right)x + 1 – \sqrt 3 = 0 \cr
& \Delta = {\left( {\sqrt 3 – 1} \right)^2} – 8\sqrt 3 \left( {1 – \sqrt 3 } \right) \cr
& = 15 – 2.5.\sqrt 3 + 3 = {\left( {5 – \sqrt 3 } \right)^2} \cr
& \sqrt \Delta = \sqrt {{{\left( {5 – \sqrt 3 } \right)}^2}} = 5 – \sqrt 3 \cr
& \Rightarrow {x_1} = {{\sqrt 3 – 1 + 5 – \sqrt 3 } \over {4\sqrt 3 }} = {{\sqrt 3 } \over 3} \cr
& {x_2} = {{\sqrt 3 – 1 – 5 + \sqrt 3 } \over {4\sqrt 3 }} = {{1 – \sqrt 3 } \over 2} \cr}\)
f)
\(\eqalign{
& {x^2} + 2\sqrt 2 x + 4 = 3\left( {x + \sqrt 2 } \right) \cr
& \Leftrightarrow {x^2} + \left( {2\sqrt 2 – 3} \right)x + 4 – 3\sqrt 2 = 0 \cr
& \Delta = 8 – 12\sqrt 2 + 9 – 16 + 12\sqrt 2 = 1 \cr
& \sqrt \Delta = 1 \cr
& \Rightarrow {x_1} = {{3 – 2\sqrt 2 + 1} \over 2} = 2 – \sqrt 2 \cr
& {x_2} = {{3 – 2\sqrt 2 – 1} \over 2} = 1 – \sqrt 2 \cr} \)