Предмет: Математика,
автор: илья1443
помогите плиз срочно 3
Приложения:
Ответы
Автор ответа:
0
(![( frac{3}{x+4}+ frac{6x}{ x^{2} +x-12}- frac{1}{x-3}): frac{8x-13}{ x^{2} -16}= \
=( frac{3}{x+4}- frac{1}{x-3}+ frac{6x}{ x^{2} +x-12})*frac{x^{2} -16}{ 8x-13}= \
=(frac{3*(x-3)-(x+4)+6x}{ x^{2} +x-12})*frac{x^{2} -16}{ 8x-13}= \
=frac{3x-9-x-4+6x}{ x^{2} +x-12}*frac{x^{2} -16}{ 8x-13}= \
=frac{8x-13}{ x^{2} +x-12}*frac{x^{2} -16}{ 8x-13}=frac{x^{2} -16}{ x^{2} +x-12}](https://tex.z-dn.net/?f=%28+frac%7B3%7D%7Bx%2B4%7D%2B+frac%7B6x%7D%7B+x%5E%7B2%7D+%2Bx-12%7D-++frac%7B1%7D%7Bx-3%7D%29%3A+frac%7B8x-13%7D%7B+x%5E%7B2%7D+-16%7D%3D+++%5C+%0A%3D%28+frac%7B3%7D%7Bx%2B4%7D-++frac%7B1%7D%7Bx-3%7D%2B+frac%7B6x%7D%7B+x%5E%7B2%7D+%2Bx-12%7D%29%2Afrac%7Bx%5E%7B2%7D+-16%7D%7B+8x-13%7D%3D++%5C+%0A%3D%28frac%7B3%2A%28x-3%29-%28x%2B4%29%2B6x%7D%7B+x%5E%7B2%7D+%2Bx-12%7D%29%2Afrac%7Bx%5E%7B2%7D+-16%7D%7B+8x-13%7D%3D+%5C+%0A%3Dfrac%7B3x-9-x-4%2B6x%7D%7B+x%5E%7B2%7D+%2Bx-12%7D%2Afrac%7Bx%5E%7B2%7D+-16%7D%7B+8x-13%7D%3D+%5C+%0A%3Dfrac%7B8x-13%7D%7B+x%5E%7B2%7D+%2Bx-12%7D%2Afrac%7Bx%5E%7B2%7D+-16%7D%7B+8x-13%7D%3Dfrac%7Bx%5E%7B2%7D+-16%7D%7B+x%5E%7B2%7D+%2Bx-12%7D "( frac{3}{x+4}+ frac{6x}{ x^{2} +x-12}- frac{1}{x-3}): frac{8x-13}{ x^{2} -16}= \
=( frac{3}{x+4}- frac{1}{x-3}+ frac{6x}{ x^{2} +x-12})*frac{x^{2} -16}{ 8x-13}= \
=(frac{3*(x-3)-(x+4)+6x}{ x^{2} +x-12})*frac{x^{2} -16}{ 8x-13}= \
=frac{3x-9-x-4+6x}{ x^{2} +x-12}*frac{x^{2} -16}{ 8x-13}= \
=frac{8x-13}{ x^{2} +x-12}*frac{x^{2} -16}{ 8x-13}=frac{x^{2} -16}{ x^{2} +x-12}")
)
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