Question

Aduceti la o forma simpla expresia:b si c​

Answer 1

Răspuns:

b)

$(\frac{1+x}{1-x} -\frac{4x}{1-x^{2} } ):\frac{1-x}{x} -\frac{1}{x+1} =(\frac{1+x)1+x}{1-x} -\frac{4x}{(1-x)(1+x)}*\frac{x}{1-x}-\frac{1}{x+1} =\frac{1+2x+x^{2}-4x}{(1-x)(1+x)} *\frac{x}{1-x}-\frac{1}{x+1} =\frac{1-2x+x^{2}}{(1-x)(1+x)} *\frac{x}{1-x}-\frac{1}{x+1} =\frac{(1-x)^{2}}{(1-x)(1+x)} *\frac{x}{1-x}-\frac{1}{x+1} =\frac{x}{1+x} -\frac{1}{x+1} =\frac{x-1}{1+x}$

c)

$\frac{x^{2}-8x+15}{x^{2}-5x+6}:\frac{x^{2}-9x+20}{x^{2}-4x+3}=\frac{x^{2}-8x+16-1}{x^{2}-4x+4-x+2}*\frac{x^{2}-4x+4-1}{x^{2}-8x+16-4+4}=\frac{(x-4)^{2}-1}{(x-2)^{2}-(x-2)}*\frac{(x-2)^{2}-1}{(x-4)^{2}-(x-4)}=\frac{(x-5)(x-3)}{(x-2)(x-3)}*\frac{(x-3)(x-1)}{(x-4)(x-5)}=\frac{x-5}{x-2}=\frac{(x-3)(x-1)}{(x-4)(x-3)}=\frac{(x-3)(x-1)}{x-2}$

Explicație pas cu pas:

Am folosit formulele de calcul prescurtat:

$(a+b)^{2} =a^{2}+2ab+b^{2}$

$(a-b)^{2} =a^{2}-2ab+b^{2}$

$a^{2} -b^{2} =(a+b)(a-b)$