Câu 44: Biến đổi các biểu thức sau thành phân thức
a. ({1 over 2} + {x over {1 – {x over {x + 2}}}})
b. ({{x – {1 over {{x^2}}}} over {x + {1 over x} + {1 over {{x^2}}}}})
c. ({{1 – {{2y} over x} + {{{y^2}} over {{x^2}}}} over {{1 over x} – {1 over y}}})
d. ({{{x over 4} – 1 + {3 over {4x}}} over {{x over 2} – {6 over x} + {1 over 2}}})
a. ({1 over 2} + {x over {1 – {x over {x + 2}}}})( = {1 over 2} + {x over {{{x + 2 – x} over {x + 2}}}} = {1 over 2} + {x over {{2 over {x + 2}}}})
b. ({{x – {1 over {{x^2}}}} over {x + {1 over x} + {1 over {{x^2}}}}}) ( = left( {x – {1 over {{x^2}}}} right):left( {1 + {1 over x} + {1 over {{x^2}}}} right) = {{{x^3} – 1} over {{x^2}}}:{{{x^2} + x + 1} over {{x^2}}})
( = {{{x^3} – 1} over {{x^2}}}.{{{x^2}} over {{x^2} + x + 1}} = {{left( {x – 1} right)left( {{x^2} + x + 1} right){x^2}} over {{x^2}left( {{x^2} + x + 1} right)}} = x – 1)
c. ({{1 – {{2y} over x} + {{{y^2}} over {{x^2}}}} over {{1 over x} – {1 over y}}})( = left( {1 – {{2y} over x} + {{{y^2}} over {{x^2}}}} right):left( {{1 over x} – {1 over y}} right) = {{{x^2} – 2xy + {y^2}} over {{x^2}}}:{{y – x} over {xy}})
( = {{{x^2} – 2xy + {y^2}} over {{x^2}}}.{{xy} over {y – x}} = {{{{left( {y – x} right)}^2}.xy} over {{x^2}left( {y – x} right)}} = {{yleft( {y – x} right)} over x})
d. ({{{x over 4} – 1 + {3 over {4x}}} over {{x over 2} – {6 over x} + {1 over 2}}})( = left( {{x over 4} – 1 + {3 over {4x}}} right):left( {{x over 2} – {6 over x} + {1 over 2}} right) = {{{x^2} – 4x + 3} over {4x}}:{{{x^2} – 12x + x} over {2x}})
(eqalign{ & = {{{x^2} – 4x + 3} over {4x}}.{{2x} over {{x^2} – 12 + x}} = {{{x^2} – x – 3x + 3} over {4x}}.{{2x} over {{x^2} – 3x + 4x – 12}} cr & = {{left( {x – 1} right)left( {x – 3} right)} over {4x}}.{{2x} over {left( {x – 3} right)left( {x + 4} right)}} = {{left( {x – 1} right)left( {x – 3} right).2x} over {4xleft( {x – 3} right)left( {x + 4} right)}} = {{x – 1} over {2left( {x + 4} right)}} cr} )
Câu 45: Thực hiện các phép tính sau :
a. (left( {{{5x + y} over {{x^2} – 5xy}} + {{5x – y} over {{x^2} + 5xy}}} right).{{{x^2} – 25{y^2}} over {{x^2} + {y^2}}})
b. ({{4xy} over {{y^2} – {x^2}}}:left( {{1 over {{x^2} + 2xy + {y^2}}} – {1 over {{x^2} – {y^2}}}} right))
c. (left[ {{1 over {{{left( {2x – y} right)}^2}}} + {2 over {4{x^2} – {y^2}}} + {1 over {{{left( {2x + y} right)}^2}}}} right].{{4{x^2} + 4xy + {y^2}} over {16x}})
d. (left( {{2 over {x + 2}} – {4 over {{x^2} + 4x + 4}}} right):left( {{2 over {{x^2} – 4}} + {1 over {2 – x}}} right))
a. (left( {{{5x + y} over {{x^2} – 5xy}} + {{5x – y} over {{x^2} + 5xy}}} right).{{{x^2} – 25{y^2}} over {{x^2} + {y^2}}})
(eqalign{ & = left[ {{{5x + y} over {xleft( {x – 5y} right)}} + {{5x – y} over {xleft( {x + 5y} right)}}} right].{{{x^2} – 25{y^2}} over {{x^2} + {y^2}}} cr & = {{left( {5x + y} right)left( {x + 5y} right) + left( {5x – y} right)left( {x – 5y} right)} over {xleft( {x – 5y} right)left( {x + 5y} right)}}.{{left( {x – 5y} right)left( {x + 5y} right)} over {{x^2} + {y^2}}} cr & = {{5{x^2} + 25xy + xy + 5{y^2} + 5{x^2} – 25xy – xy + 5{y^2}} over {xleft( {{x^2} + {y^2}} right)}} cr & = {{10{x^2} + 10{y^2}} over {xleft( {{x^2} + {y^2}} right)}} = {{10left( {{x^2} + {y^2}} right)} over {xleft( {{x^2} + {y^2}} right)}} = {{10} over x} cr} )
b. ({{4xy} over {{y^2} – {x^2}}}:left( {{1 over {{x^2} + 2xy + {y^2}}} – {1 over {{x^2} – {y^2}}}} right))
(eqalign{ & = {{4xy} over {{y^2} – {x^2}}}:left[ {{1 over {{{left( {x + y} right)}^2}}} – {1 over {left( {x + y} right)left( {x – y} right)}}} right] cr & = {{4xy} over {{y^2} – {x^2}}}:{{x – y – left( {x + y} right)} over {{{left( {x + y} right)}^2}left( {x – y} right)}} = {{4xy} over {{y^2} – {x^2}}}:{{ – 2y} over {{{left( {x + y} right)}^2}left( {x – y} right)}} = {{4xy} over {{y^2} – {x^2}}}.{{{{left( {x + y} right)}^2}left( {y – x} right)} over {2y}} cr & = {{4xy{{left( {x + y} right)}^2}left( {y – x} right)} over {left( {y + x} right)left( {y – x} right).2y}} = 2xleft( {x + y} right) cr} )
c. (left[ {{1 over {{{left( {2x – y} right)}^2}}} + {2 over {4{x^2} – {y^2}}} + {1 over {{{left( {2x + y} right)}^2}}}} right].{{4{x^2} + 4xy + {y^2}} over {16x}})
(eqalign{ & = left[ {{1 over {{{left( {2x – y} right)}^2}}} + {2 over {left( {2x + y} right)left( {2x – y} right)}} + {1 over {{{left( {2x + y} right)}^2}}}} right].{{{{left( {2x + y} right)}^2}} over {16x}} cr & = {{{{left( {2x + y} right)}^2} + 2left( {2x + y} right)left( {2x – y} right) + {{left( {2x – y} right)}^2}} over {{{left( {2x + y} right)}^2}.{{left( {2x – y} right)}^2}}}.{{{{left( {2x + y} right)}^2}} over {16x}} cr & = {{{{left[ {left( {2x + y} right) + left( {2x – y} right)} right]}^2}} over {16x{{left( {2x – y} right)}^2}}} = {{{{left( {4x} right)}^2}} over {16x{{left( {2x – y} right)}^2}}} = {{16{x^2}} over {16x{{left( {2x – y} right)}^2}}} = {x over {{{left( {2x – y} right)}^2}}} cr} )
d. (left( {{2 over {x + 2}} – {4 over {{x^2} + 4x + 4}}} right):left( {{2 over {{x^2} – 4}} + {1 over {2 – x}}} right))
(eqalign{ & = left[ {{2 over {x + 2}} – {4 over {{{left( {x + 2} right)}^2}}}} right]:left[ {{2 over {left( {x + 2} right)left( {x – 2} right)}} – {1 over {x – 2}}} right] cr & = {{2left( {x + 2} right) – 4} over {{{left( {x + 2} right)}^2}}}:{{2 – left( {x + 2} right)} over {left( {x + 2} right)left( {x – 2} right)}} = {{2x + 4 – 4} over {{{left( {x + 2} right)}^2}}}:{{2 – x – 2} over {left( {x + 2} right)left( {x – 2} right)}} cr & = {{2x} over {{{left( {x + 2} right)}^2}}}.{{left( {x + 2} right)left( {x – 2} right)} over { – x}} = {{2left( {x – 2} right)} over { – left( {x + 2} right)}} = {{2left( {2 – x} right)} over {x + 2}} cr} )
Câu 46: Tìm điều kiện của biến để giá trị của phân thức xác định :
a. ({{5{x^2} – 4x + 2} over {20}})
b. ({8 over {x + 2004}})
c. ({{4x} over {3x – 7}})
d. ({{{x^2}} over {x + z}})
a. Phân thức : ({{5{x^2} – 4x + 2} over {20}})xác định với mọi (x in R)
b. Phân thức : ({8 over {x + 2004}})xác định khi (x + 2004 ne 0 Rightarrow x ne – 2004)
c. Phân thức : ({{4x} over {3x – 7}})xác định khi (3x – 7 ne 0 Rightarrow x ne {7 over 3})
d. Phân thức : ({{{x^2}} over {x + z}})xác định khi (x + z ne 0 Rightarrow x ne – z)
Câu 47: Phân tích mẫu thức của các phân thức sau thành nhân tử rồi tìm điều kiện của x để giá trị của phân thức xác định :
a. ({5 over {2x – 3{x^2}}})
b. ({{2x} over {8{x^3} + 12{x^2} + 6x + 1}})
c. ({{ – 5{x^2}} over {16 – 24x + 9{x^2}}})
d. ({3 over {{x^2} – 4{y^2}}})
a. ({5 over {2x – 3{x^2}}})( = {5 over {xleft( {2 – 3x} right)}}) xác định khi (xleft( {2 – 3x} right) ne 0)
(left{ {matrix{{x ne 0} cr{2 – 3x ne 0} cr} Rightarrow left{ {matrix{ {x ne 0} cr {x ne {2 over 3}} cr} } right.} right.)
Vậy phân thức ({5 over {2x – 3{x^2}}}) xác định với (x ne 0) và (x ne {2 over 3})
b. ({{2x} over {8{x^3} + 12{x^2} + 6x + 1}}) ( = {{2x} over {{{left( {2x + 1} right)}^3}}}) xác định khi ({left( {2x + 1} right)^3} ne 0 Rightarrow 2x + 1 ne 0 Rightarrow x ne – {1 over 2})
c. ({{ – 5{x^2}} over {16 – 24x + 9{x^2}}})( = {{ – 5{x^2}} over {{4^2} – 2.4.3x + {{left( {3x} right)}^2}}} = {{ – 5{x^2}} over {{{left( {4 – 3x} right)}^2}}})
xác định khi ({left( {4 – 3x} right)^2} ne 0 Rightarrow 4 – 3x ne 0 Rightarrow x ne {4 over 3})
d. ({3 over {{x^2} – 4{y^2}}})( = {3 over {left( {x – 2y} right)left( {x + 2y} right)}}) xác định khi (left( {x – 2y} right)left( {x + 2y} right) ne 0)
( Rightarrow left{ {matrix{{x – 2y ne 0} cr{x + 2y ne 0} cr} Rightarrow x ne pm 2y} right.)
