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loshadbegi
@loshadbegi
August 2022
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Помогите с алгеброй плиз, дам 60 баллов!
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patritsiagonchar
Ответ:
Объяснение:
1. а)
б)
2. a)
б)
2 votes
Thanks 1
loshadbegi
2 задание?
patritsiagonchar
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Answers & Comments
Ответ:
Объяснение:
1. а)![\frac{6a^{3} b - 6ab^{3} }{a^{4} - b^{4} } =\frac{6ab * (a^{2} - b^{2} }{(a^{2}-b^{2} )*(a^{2} + b^{2}) } =\frac{6ab}{a^{2}+b^{2} } \frac{6a^{3} b - 6ab^{3} }{a^{4} - b^{4} } =\frac{6ab * (a^{2} - b^{2} }{(a^{2}-b^{2} )*(a^{2} + b^{2}) } =\frac{6ab}{a^{2}+b^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B6a%5E%7B3%7D%20b%20-%206ab%5E%7B3%7D%20%7D%7Ba%5E%7B4%7D%20-%20b%5E%7B4%7D%20%7D%20%3D%5Cfrac%7B6ab%20%2A%20%28a%5E%7B2%7D%20-%20b%5E%7B2%7D%20%7D%7B%28a%5E%7B2%7D-b%5E%7B2%7D%20%29%2A%28a%5E%7B2%7D%20%2B%20b%5E%7B2%7D%29%20%7D%20%3D%5Cfrac%7B6ab%7D%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%20%20%7D)
б)![\frac{9x+6x+1}{9x^{2}-1 } =\frac{(3x+1)^{2}}{ (3x-1)*(3x+1) } =\frac{3x+1}{3x-1} \frac{9x+6x+1}{9x^{2}-1 } =\frac{(3x+1)^{2}}{ (3x-1)*(3x+1) } =\frac{3x+1}{3x-1}](https://tex.z-dn.net/?f=%5Cfrac%7B9x%2B6x%2B1%7D%7B9x%5E%7B2%7D-1%20%7D%20%3D%5Cfrac%7B%283x%2B1%29%5E%7B2%7D%7D%7B%20%283x-1%29%2A%283x%2B1%29%20%7D%20%3D%5Cfrac%7B3x%2B1%7D%7B3x-1%7D)
2. a)![\frac{2a}{c^{3} } +\frac{3}{c^{2} } =\frac{2a}{c^{3} } +\frac{c*3}{c*c^{2} } =\frac{2a}{c^{3} } +\frac{3c}{c^{3} } = \frac{2a+3c}{c^{3} } \frac{2a}{c^{3} } +\frac{3}{c^{2} } =\frac{2a}{c^{3} } +\frac{c*3}{c*c^{2} } =\frac{2a}{c^{3} } +\frac{3c}{c^{3} } = \frac{2a+3c}{c^{3} }](https://tex.z-dn.net/?f=%5Cfrac%7B2a%7D%7Bc%5E%7B3%7D%20%7D%20%2B%5Cfrac%7B3%7D%7Bc%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B2a%7D%7Bc%5E%7B3%7D%20%7D%20%2B%5Cfrac%7Bc%2A3%7D%7Bc%2Ac%5E%7B2%7D%20%7D%20%3D%5Cfrac%7B2a%7D%7Bc%5E%7B3%7D%20%7D%20%2B%5Cfrac%7B3c%7D%7Bc%5E%7B3%7D%20%7D%20%3D%20%5Cfrac%7B2a%2B3c%7D%7Bc%5E%7B3%7D%20%7D)
б)![\frac{a}{3-4a}-\frac{a^{3} }{16a^{2} -9} = \frac{a}{-(4a-3)}- \frac{a^{3} }{(4a-3)*(4a+3)} =-\frac{a}{4a-3} -\frac{a^{3} }{(4a-3)*(4a+3)} =-\frac{a*(4a+3)+a^{3}}{(4a-3*(4a+3)} = \frac{a}{3-4a}-\frac{a^{3} }{16a^{2} -9} = \frac{a}{-(4a-3)}- \frac{a^{3} }{(4a-3)*(4a+3)} =-\frac{a}{4a-3} -\frac{a^{3} }{(4a-3)*(4a+3)} =-\frac{a*(4a+3)+a^{3}}{(4a-3*(4a+3)} =](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7B3-4a%7D-%5Cfrac%7Ba%5E%7B3%7D%20%7D%7B16a%5E%7B2%7D%20-9%7D%20%20%3D%20%5Cfrac%7Ba%7D%7B-%284a-3%29%7D-%20%5Cfrac%7Ba%5E%7B3%7D%20%7D%7B%284a-3%29%2A%284a%2B3%29%7D%20%3D-%5Cfrac%7Ba%7D%7B4a-3%7D%20-%5Cfrac%7Ba%5E%7B3%7D%20%7D%7B%284a-3%29%2A%284a%2B3%29%7D%20%3D-%5Cfrac%7Ba%2A%284a%2B3%29%2Ba%5E%7B3%7D%7D%7B%284a-3%2A%284a%2B3%29%7D%20%3D)
![= \frac{4a^{2}+3a+a^{3} }{16a^{2}-9 } = \frac{4a^{2}+3a+a^{3} }{16a^{2}-9 }](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4a%5E%7B2%7D%2B3a%2Ba%5E%7B3%7D%20%20%7D%7B16a%5E%7B2%7D-9%20%7D)