Nội dung bài giảng
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)
\(\eqalign{
& {{{x^2}} \over 5} - {{2{\rm{x}}} \over 3} = {{x + 5} \over 6} \cr
& \Leftrightarrow 6{{\rm{x}}^2} - 20{\rm{x}} = 5{\rm{x}} + 25 \cr
& \Leftrightarrow 6{{\rm{x}}^2} - 25{\rm{x}} - 25 = 0 \cr
& \Delta = {25^2} + 4.6.25 = 1225 \cr
& \sqrt \Delta = 35 \Rightarrow {x_1} = 5;{x_2} = - {5 \over 6} \cr} \)
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)
\(\eqalign{
& {x^2} + 2\sqrt 2 x + 4 = 3\left( {x + \sqrt 2 } \right) \cr
& \Leftrightarrow {x^2} + \left( {2\sqrt 2 - 3} \right)x + 4 - 3\sqrt 2 = 0 \cr
& \Delta = 8 - 12\sqrt 2 + 9 - 16 + 12\sqrt 2 = 1 \cr
& \sqrt \Delta = 1 \cr
& \Rightarrow {x_1} = {{3 - 2\sqrt 2 + 1} \over 2} = 2 - \sqrt 2 \cr
& {x_2} = {{3 - 2\sqrt 2 - 1} \over 2} = 1 - \sqrt 2 \cr} \)