Контролно по математика - алгебра

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1. група
Задача 1.
[tex]\frac {9}{2 \sqrt {5} + \sqrt {11}}- \frac {21}{\sqrt {11}-5}- \frac {\sqrt {11} +15}{2}=[/tex]
[tex]=\frac {9}{2 \sqrt {5} + \sqrt {11}}. \frac {2 \sqrt {5} - \sqrt {11}}{2 \sqrt {5} - \sqrt {11}}+ \frac {21}{5-\sqrt {11}}. \frac {5+\sqrt {11}}{5+\sqrt {11}}- \frac {\sqrt {11} +15}{2}[/tex]
[tex]=\frac {18 \sqrt {5} - 18 \sqrt {11}}{(2 \sqrt {5})^2 -( \sqrt {11})^2}+ \frac {105+21\sqrt {11}}{5^2-(\sqrt {11})^2}- \frac {\sqrt {11} +15}{2}=[/tex]
[tex]=\frac {18 \sqrt {5} - 18 \sqrt {11}}{9}+ \frac {105+21\sqrt {11}}{14}- \frac {\sqrt {11} +15}{2}=[/tex]
Определяме най-малкото общо кратно на трите знаменателя 9; 14; 2, а именно 126.
[tex]=\frac {18 \sqrt {5} - 18 \sqrt {11}}{9}. \frac {14}{14}+ \frac {105+21\sqrt {11}}{14}. \frac {9}{9}- \frac {\sqrt {11} +15}{2}. \frac {63}{63}=[/tex]
[tex]=\frac {14 (18 \sqrt {5} - 18 \sqrt {11})+9 (105+21\sqrt {11})-63(\sqrt {11} +15)}{126}=[/tex]
[tex]=\frac {18.14 ( \sqrt {5} - \sqrt {11})+9 .21(5+\sqrt {11})-9.7(\sqrt {11} +15)}{9.14}=[/tex]
[tex]=\frac { \cancel {9} [2.14 ( \sqrt {5} - \sqrt {11})+21(5+\sqrt {11})-7(\sqrt {11} +15)]}{\cancel {9}.14}=[/tex]
[tex]=\frac { 28 \sqrt {5} - 28\sqrt {11}\cancel {+105}+ 21\sqrt {11}-7\sqrt {11} \cancel {-105}}{14}=[/tex]
[tex]=\frac { 28 \sqrt {5} - 14\sqrt {11}}{14}=\frac {\cancel {14} ( 2 \sqrt {5} - \sqrt {11})}{\cancel {14}}=[/tex]
[tex]=2 \sqrt {5} - \sqrt {11}[/tex]

Задача 2. Да се изчисли при [tex]x=2 \frac {3}{5}[/tex]
[tex](x+1 + \frac {1}{x-1}): (x+ \frac {x^2}{1-x})=[/tex]
[tex]=( \frac {x+1}{1} + \frac {1}{x-1}): (\frac {x}{1}+ \frac {x^2}{1-x})=[/tex]
[tex]= \frac {(x+1)(x-1)+1}{x-1}: \frac {x(1-x)+x^2}{1-x}=[/tex]
[tex]= - \frac {x^2 \cancel {-1} \cancel {+1}}{\cancel {1-x}}. \frac{\cancel {1-x}} {x\cancel{-x^2}\cancel{+x^2}}= \frac {x^{\cancel {2}}}{\cancel {x}}= x= 2 \frac35[/tex]

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Задача 3. Да се докаже тъждеството:
[tex]( \frac{a^2+2b^2}{ab} - \frac{a^2}{ab-b^2} + \frac{b^2}{a^2-ab} ). \frac{a}{b-a} =1[/tex]
[tex]( \frac{a^2+2b^2}{ab} - \frac{a^2}{b(a-b)} + \frac{b^2}{a(a-b)} ). \frac{a}{b-a} =[/tex]
След намиране на най-малкото общо кратно, а именно [tex]ab(a-b)[/tex], разширяваме дробите
[tex]=( \frac{(a^2+2b^2)(a-b)}{ab(a-b)} - \frac{a^2.a}{ab(a-b)} + \frac{b^2.b}{ab(a-b)} ). \frac{a}{b-a} =[/tex]
[tex]= \frac{(a^2+2b^2)(a-b)-a^3+b^3}{\cancel {a}b(a-b)} . \frac{\cancel {a}}{b-a} =[/tex]
[tex]= \frac{\cancel {a^3}+2ab^2-a^2b-2b^3 \cancel {-a^3}+b^3}{b(a-b)} . \frac{1}{b-a} =[/tex]
[tex]= \frac{\cancel {b} (2ab-a^2-b^2)}{\cancel {b}(a-b)} . \frac{1}{b-a} =[/tex]
[tex]= \frac{-( 2ab+a^2+b^2)}{-(b-a)(b-a)} =\frac {b^2-2ab+a^2}{(b-a)^2}=\frac {\cancel {(b-a)^2}}{\cancel {(b-a)^2}}=1[/tex]
Извод: равенството е тъждество

Задача 4. Да се пресметне за [tex]a=9; b=20[/tex]
[tex](\sqrt {a} + \frac {b- \sqrt {ab}}{\sqrt {a}+ \sqrt {b}}):(\frac{a}{\sqrt {ab}+b}+\frac{b}{\sqrt {ab}-a}-\frac{a+b}{\sqrt {ab}})=[/tex]
[tex]=(\frac {\sqrt {a}}{1} + \frac {b- \sqrt {ab}}{\sqrt {a}+ \sqrt {b}}):(\frac{a}{\sqrt {ab}+b}.\frac{\sqrt{ab}(\sqrt {ab}-a)}{\sqrt{ab}(\sqrt {ab}-a)}+\frac{b}{\sqrt {ab}-a}.\frac{\sqrt{ab}(\sqrt {ab}+b)}{\sqrt{ab}(\sqrt {ab}+b)}-\frac{a+b}{\sqrt {ab}}. \frac {(\sqrt {ab}-a)(\sqrt {ab}+b)}{(\sqrt {ab}-a)(\sqrt {ab}+b)})[/tex]
[tex]=\frac {a \cancel {+\sqrt {ab} } + b \cancel {- \sqrt {ab} }}{\sqrt {a}+ \sqrt {b}}:\frac{a. \sqrt{ab}(\sqrt {ab}-a)+b. \sqrt{ab}(\sqrt {ab}+b) - (a+b) (\sqrt {ab}-a)(\sqrt {ab}+b)}{\sqrt{ab}(\sqrt {ab}-a) (\sqrt {ab}+b)}=[/tex]
[tex]=\frac {a + b }{\sqrt {a}+ \sqrt {b}}:\frac{a^2b-a^2 \sqrt {ab}+ab^2+b^2 \sqrt {ab} - (a+b) (\cancel {ab}+b \sqrt {ab}-a\sqrt {ab} \cancel {- ab})}{\sqrt{ab}(\cancel {ab}+b\sqrt {ab} -a \sqrt {ab} \cancel { - ab})}=[/tex]
[tex]=\frac {a + b }{\sqrt {a}+ \sqrt {b}}:\frac{a^2b-a^2 \sqrt {ab}+ab^2+b^2 \sqrt {ab} - (a+b) (b \sqrt {ab}-a\sqrt {ab} )}{\sqrt{ab}(b\sqrt {ab} -a \sqrt {ab})}=[/tex]
[tex]=\frac {a + b }{\sqrt {a}+ \sqrt {b}}:\frac{a^2b \cancel {-a^2 \sqrt {ab}}+ab^2 \cancel {+b^2 \sqrt {ab}} \cancel {- ab \sqrt {ab}} \cancel {+ a^2 \sqrt {ab}} \cancel {-b^2 \sqrt {ab}} \cancel {+ab \sqrt {ab}}}{\sqrt{ab}.\sqrt {ab}(b-a)}=[/tex]
[tex]=\frac {a + b }{\sqrt {a}+ \sqrt {b}}. \frac {\sqrt{ab}.\sqrt {ab}(b-a)} {a^2b +ab^2}=\frac { \cancel {a + b }}{\sqrt {a}+ \sqrt {b}}. \frac {\cancel {ab}(b-a)} {\cancel {ab} \cancel {(a+b)}}=\frac {b-a}{ \sqrt {b}+\sqrt {a}}. \frac {\sqrt {b}-\sqrt {a}}{\sqrt {b}-\sqrt {a}}=[/tex]
[tex]=\frac { \cancel {a + b }}{\sqrt {a}+ \sqrt {b}}. \frac {\cancel {ab}(b-a)} {\cancel {ab} \cancel {(a+b)}}=\frac {\cancel {b-a}}{\cancel {b-a}}. (\sqrt {b}-\sqrt {a})[/tex]
[tex]\sqrt {b}-\sqrt {a}=\sqrt {20}- \sqrt{9} = 2\sqrt {5} - 3[/tex]

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1. група
Задача 1.
[tex]\frac {9}{2 \sqrt {5} + \sqrt {11}}- \frac {2}{2 \sqrt {2}-3 \sqrt {3}}=[/tex]
[tex]= \frac {9}{2 \sqrt {5} + \sqrt {11}}. \frac {2 \sqrt {5} - \sqrt {11}}{2 \sqrt {5} - \sqrt {11}}- \frac {2}{2 \sqrt {2}-3 \sqrt {3}}.\frac {2 \sqrt {2}+3 \sqrt {3}}{2 \sqrt {2}+3 \sqrt {3}}=[/tex]
[tex]= \frac {9(2 \sqrt {5} - \sqrt {11})}{(2 \sqrt {5})^2 -( \sqrt {11})^2}- \frac {2(2 \sqrt {2}+3 \sqrt {3})}{(2 \sqrt {2})^2-(3 \sqrt {3})^2}=[/tex]
[tex]= \frac {18 \sqrt {5} - 9 \sqrt {11}}{20-11}- \frac {4 \sqrt {2}+6 \sqrt {3}}{8-27}=[/tex]
[tex]= \frac {18 \sqrt {5} - 9 \sqrt {11}}{9}+ \frac {4 \sqrt {2}+6 \sqrt {3}}{19}=[/tex]
[tex]= \frac {18 \sqrt {5} - 9 \sqrt {11}}{9}.\frac {19}{19}+ \frac {4 \sqrt {2}+6 \sqrt {3}}{19}. \frac {9}{9}=[/tex]
[tex]= \frac {342 \sqrt {5} - 171 \sqrt {11}+36 \sqrt {2}+54 \sqrt {3}}{9.19}=[/tex]
[tex]= \frac {\cancel {9}(38 \sqrt {5} - 19 \sqrt {11}+4 \sqrt {2}+6 \sqrt {3})}{\cancel {9}.19}=[/tex]
[tex]= \frac {38 \sqrt {5} - 19 \sqrt {11}+4 \sqrt {2}+6 \sqrt {3}}{19}[/tex]

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Задача 2. Пресметнете при [tex]x= 1 \frac {12}{13}[/tex]
[tex](\frac {x}{x+2}-{x-2}{x^2+2x}): \frac {x^2-x+2}{x+2}=[/tex]
[tex]=(\frac {x}{x+2}.\frac {x}{x}-\frac{x-2}{x(x+2)}). \frac {x+2}{x^2-x+2}=[/tex]
[tex]=\frac{\cancel {x^2-x+2}}{x\cancel {(x+2)}}. \frac {\cancel {x+2}}{\cancel {x^2-x+2}}=[/tex]
[tex]\frac {1}{x}= \frac {1}{1 \frac {12}{13}}1 = \frac {1}{\frac {13.1+12}{13}}= \frac {1}{\frac {25}{13}} = \frac {13}{25}[/tex]

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Задача 3. Докажете тъждеството
[tex](\frac{b-a}{a}+1-b).\frac {a+b}{b-ba^2}.\frac {5+5a}{ab+b^2}= \frac {5}{ab}[/tex]
[tex](\frac{b-a}{a}+\frac{1-b}{1}. \frac {a}{a}).\frac {a+b}{b(1-a^2)}.\frac {5+5a}{b(a+b)}=[/tex]
[tex]=\frac{b\cancel {-a} \cancel {+a}-ab}{a}.\frac {a+b}{b(1-a)(1+a)}.\frac {5(1+a)}{b(a+b)}=[/tex]
[tex]=\frac{\cancel {b}\cancel {(1-a)}}{a}.\frac {\cancel {a+b}}{b\cancel {(1-a)}\cancel {(1+a)}}.\frac {5\cancel {(1+a)}}{\cancel {b}\cancel {(a+b)}}= \frac {5}{ab}[/tex]
Равенството е тъждество

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Задача 4. Пресметнете при [tex]a=1,21[/tex]
[tex](\frac {\sqrt {a}}{2}- \frac {1}{2 \sqrt {a}})^2.(\frac {\sqrt {a}-1}{\sqrt {a}+1}- \frac {\sqrt {a}+1}{\sqrt {a}-1})=[/tex]
В скобите намираме НОК и разширяваме дробите
[tex]=(\frac {\sqrt {a}}{2}. \frac {\sqrt {a}}{\sqrt {a}} - \frac {1}{2 \sqrt {a}})^2.(\frac {\sqrt {a}-1}{\sqrt {a}+1}. \frac {\sqrt {a}-1}{\sqrt {a}-1}- \frac {\sqrt {a}+1}{\sqrt {a}-1}. \frac {\sqrt {a}+1}{\sqrt {a}+1} )=[/tex]

[tex]=(\frac {a-1}{2 \sqrt {a}})^2.\frac {(\sqrt {a}-1)^2-(\sqrt {a}+1)^2}{(\sqrt {a})^2-1^2}=[/tex]
[tex]=\frac {(a-1)^{\cancel {2}}}{4a}.\frac { \cancel { a}-2 \sqrt {a} \cancel {+ 1 } \cancel {- a} - 2 \sqrt {a}\cancel {- 1}}{\cancel {a-1}}=[/tex]
[tex]= \frac {-\cancel {4} \sqrt {a} (a-1)}{\cancel {4}a}=\frac {\sqrt {a} (1-a)}{a}[/tex]
[tex]\frac {\sqrt {a} (1-a)}{a}=\frac {\sqrt {1,21} (1-1,21)}{1,21}=\frac {1,1 .(-0,21)}{1,21}=\frac {\cancel {1,1} .(-0,21)}{1,1\cancel {.1,1}}= - \frac {21}{110}[/tex]