Đề bài
Tính:
a) (left( {a + 1 + frac{{1 - 2{a^2}}}{{a - 1}}} right):left( {1 - frac{1}{{1 - a}}} right));
b) (left( {frac{a}{{{b^2}}} - frac{1}{a}} right):left( {frac{1}{b} + frac{1}{a}} right));
c) (left( {a - frac{{4ab}}{{a + b}} + b} right).left( {a + frac{{4ab}}{{a - b}} - b} right));
d) (ab + frac{{ab}}{{a + b}}left( {frac{{a + b}}{{a - b}} - a - b} right)).
Lời giải chi tiết
a) (left( {a + 1 + frac{{1 - 2{a^2}}}{{a - 1}}} right):left( {1 - frac{1}{{1 - a}}} right) = frac{{left( {a + 1} right)left( {a - 1} right) + 1 - 2{a^2}}}{{a - 1}}:frac{{1 - a - 1}}{{1 - a}})
( = frac{{{a^2} - 1 + 1 - 2{a^2}}}{{a - 1}}.frac{{a - 1}}{a} = frac{{ - {a^2}left( {a - 1} right)}}{{aleft( {a - 1} right)}} = - a)
b) (left( {frac{a}{{{b^2}}} - frac{1}{a}} right):left( {frac{1}{b} + frac{1}{a}} right) = frac{{{a^2} - {b^2}}}{{a{b^2}}}:frac{{a + b}}{{ab}} = frac{{left( {a - b} right)left( {a + b} right)ab}}{{a{b^2}left( {a + b} right)}} = frac{{a - b}}{b})
c) (left( {a - frac{{4ab}}{{a + b}} + b} right).left( {a + frac{{4ab}}{{a - b}} - b} right) = frac{{left( {a + b} right)left( {a + b} right) - 4ab}}{{a + b}}.frac{{left( {a - b} right)left( {a - b} right) + 4ab}}{{a - b}})
( = frac{{{a^2} + 2ab + {b^2} - 4ab}}{{a + b}}.frac{{{a^2} - 2ab + {b^2} + 4ab}}{{a - b}} = frac{{{a^2} - 2ab + {b^2}}}{{a + b}}.frac{{{a^2} + 2ab + {b^2}}}{{a - b}})
( = frac{{{{left( {a - b} right)}^2}{{left( {a + b} right)}^2}}}{{left( {a + b} right)left( {a - b} right)}} = left( {a + b} right)left( {a - b} right) = {a^2} - {b^2})
d) (ab + frac{{ab}}{{a + b}}left( {frac{{a + b}}{{a - b}} - a - b} right) = ab + frac{{ab}}{{a + b}}.frac{{a + b - left( {a - b} right)left( {a + b} right)}}{{a - b}})
( = ab + frac{{ab}}{{a + b}}.frac{{left( {a + b} right)left( {1 - a + b} right)}}{{a - b}} = ab + frac{{ableft( {1 - a + b} right)}}{{a - b}} = frac{{ableft( {a - b} right) + ab - {a^2}b + a{b^2}}}{{a - b}})
( = frac{{{a^2}b - a{b^2} + ab - {a^2}b + a{b^2}}}{{a - b}} = frac{{left( {{a^2}b - {a^2}b} right) + left( {a{b^2} - a{b^2}} right) + ab}}{{a - b}} = frac{{ab}}{{a - b}})