Identités remarquables/Exercices/Quotient

Exercice 5-1

Simplifier les fractions :

a) ${\frac {x^{2}-1}{x^{2}-2x+1}}$

Solution

${\frac {x^{2}-1}{x^{2}-2x+1}}={\frac {(x+1)(x-1)}{(x-1)^{2}}}={\frac {x+1}{x-1}}$

b) ${\frac {x^{2}-1}{x^{2}+2x+1}}$

Solution

${\frac {x^{2}-1}{x^{2}+2x+1}}={\frac {(x+1)(x-1)}{(x+1)^{2}}}={\frac {x-1}{x+1}}$

c) ${\frac {x^{2}-4x+4}{x^{2}-4}}$

Solution

${\frac {x^{2}-4x+4}{x^{2}-4}}={\frac {x^{2}-2\times 2x+2^{2}}{x^{2}-2^{2}}}={\frac {(x-2)^{2}}{(x+2)(x-2)}}={\frac {x-2}{x+2}}$

d) ${\frac {(2x-1)^{2}-(x+1)^{2}}{x^{2}-4x+4}}$

Solution

${\frac {(2x-1)^{2}-(x+1)^{2}}{x^{2}-4x+4}}={\frac {(2x-1+x+1)(2x-1-x-1)}{x^{2}-2\times 2x+2^{2}}}={\frac {3x(x-2)}{(x-2)^{2}}}={\frac {3x}{x-2}}$

e) ${\frac {x^{2}+6x+9-(2x+5)^{2}}{(2x+3)^{2}-(x+1)^{2}}}$

Solution

${\begin{aligned}{\frac {x^{2}+6x+9-(2x+5)^{2}}{(2x+3)^{2}-(x+1)^{2}}}&={\frac {(x+3)^{2}-(2x+5)^{2}}{(2x+3)^{2}-(x+1)^{2}}}\\&={\frac {(x+3+2x+5)(x+3-2x-5)}{(2x+3+x+1)(2x+3-x-1)}}\\&={\frac {(3x+8)(-x-2)}{(3x+4)(x+2)}}\\&={\frac {-(3x+8)(x+2)}{(3x+4)(x+2)}}\\&=-{\frac {3x+8}{3x+4}}\end{aligned}}$

f) ${\frac {(3a-b)^{2}-(4a+b)^{2}}{3a^{2}b+6ab^{2}}}$

Solution

${\begin{aligned}{\frac {(3a-b)^{2}-(4a+b)^{2}}{3a^{2}b+6ab^{2}}}&={\frac {(3a-b+4a+b)(3a-b-4a-b)}{3ab(a+2b)}}\\&={\frac {7a(-a-2b)}{3ab(a+2b)}}\\&={\frac {-7a(a+2b)}{3ab(a+2b)}}\\&={\frac {-7}{3b}}\end{aligned}}$

Exercice 5-2

Effectuer les calculs :

a) ${\frac {2}{x^{2}-1}}+{\frac {3}{x+1}}$

Solution

${\frac {2}{x^{2}-1}}+{\frac {3}{x+1}}={\frac {2}{x^{2}-1}}+{\frac {3(x-1)}{(x+1)(x-1)}}={\frac {2+3(x-1)}{x^{2}-1}}={\frac {3x-1}{x^{2}-1}}$

b) ${\frac {3}{x^{2}-1}}-{\frac {9}{2(x-1)(x-2)}}$

Solution

${\begin{aligned}{\frac {3}{x^{2}-1}}-{\frac {9}{2(x-1)(x-2)}}&={\frac {3}{(x+1)(x-1)}}-{\frac {9}{2(x-1)(x-2)}}\\&={\frac {2\times 3(x-2)}{2(x+1)(x-1)(x-2)}}-{\frac {9(x+1)}{2(x-1)(x+1)(x-2)}}\\&={\frac {6(x-2)-9(x+1)}{2(x+1)(x-1)(x-2)}}\\&={\frac {-3x-21}{2(x+1)(x-1)(x-2)}}\\&={\frac {-3(x+7)}{2(x+1)(x-1)(x-2)}}\end{aligned}}$

c) ${\frac {2x+1}{2x^{2}-8x+8}}-{\frac {x+3}{x^{2}-4}}$

Solution

${\begin{aligned}{\frac {2x+1}{2x^{2}-8x+8}}-{\frac {x+3}{x^{2}-4}}&={\frac {2x+1}{2(x^{2}-4x+2^{2})}}-{\frac {x+3}{x^{2}-2^{2}}}\\&={\frac {2x+1}{2(x-2)^{2}}}-{\frac {x+3}{(x+2)(x-2)}}\\&={\frac {(2x+1)(x+2)}{2(x-2)^{2}(x+2)}}-{\frac {2(x-2)(x+3)}{2(x+2)(x-2)^{2}}}\\&={\frac {(2x+1)(x+2)-2(x-2)(x+3)}{2(x+2)(x-2)^{2}}}\\&={\frac {2x^{2}+4x+x+2-2x^{2}-6x+4x+12}{2(x+2)(x-2)^{2}}}\\&={\frac {3x+14}{2(x+2)(x-2)^{2}}}\end{aligned}}$

d) ${\frac {1}{1-x}}+{\frac {1}{1+x}}-{\frac {4x^{2}}{1-x^{4}}}$

Solution

${\begin{aligned}{\frac {1}{1-x}}+{\frac {1}{1+x}}-{\frac {4x^{2}}{1-x^{4}}}&={\frac {1+x}{(1-x)(1+x)}}+{\frac {1-x}{(1+x)(1-x)}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {1+x}{1-x^{2}}}+{\frac {1-x}{1-x^{2}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {1+x+1-x}{1-x^{2}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2}{1-x^{2}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2(1+x^{2})}{(1-x^{2})(1+x^{2})}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2+2x^{2}}{1-x^{4}}}-{\frac {4x^{2}}{1-x^{4}}}\\&={\frac {2-2x^{2}}{1-x^{4}}}\\&={\frac {2(1-x^{2})}{(1+x^{2})(1-x^{2})}}\\&={\frac {2}{1+x^{2}}}\end{aligned}}$

e) ${\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {1}{x-1}}-{\frac {1}{x+1}}$

Solution

${\begin{aligned}{\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {1}{x-1}}-{\frac {1}{x+1}}&={\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {x+1}{x^{2}-1}}-{\frac {x-1}{x^{2}-1}}\\&={\frac {x^{4}}{x^{4}-1}}-{\frac {x^{2}}{x^{2}-1}}+{\frac {2}{x^{2}-1}}\\&={\frac {x^{4}}{x^{4}-1}}+{\frac {2-x^{2}}{x^{2}-1}}\\&={\frac {x^{4}}{x^{4}-1}}+{\frac {(2-x^{2})(x^{2}+1)}{(x^{2}-1)(x^{2}+1)}}\\&={\frac {x^{4}}{x^{4}-1}}+{\frac {2x^{2}+2-x^{4}-x^{2}}{x^{4}-1}}\\&={\frac {x^{2}+2}{x^{4}-1}}\end{aligned}}$

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