计算①(x²-1)(x+1)(x^4+1)(x-1)因式分解①(a+3)(a+7)+25还有一道填空已知:x^n=2,y^n=3,那么(x²y)^2n=________
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![计算①(x²-1)(x+1)(x^4+1)(x-1)因式分解①(a+3)(a+7)+25还有一道填空已知:x^n=2,y^n=3,那么(x²y)^2n=________](/uploads/image/z/1239700-4-0.jpg?t=%E8%AE%A1%E7%AE%97%E2%91%A0%EF%BC%88x%26sup2%3B%EF%BC%8D1%EF%BC%89%EF%BC%88x%EF%BC%8B1%EF%BC%89%EF%BC%88x%5E4%EF%BC%8B1%EF%BC%89%EF%BC%88x%EF%BC%8D1%EF%BC%89%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E2%91%A0%28a%EF%BC%8B3%29%28a%EF%BC%8B7%29%EF%BC%8B25%E8%BF%98%E6%9C%89%E4%B8%80%E9%81%93%E5%A1%AB%E7%A9%BA%E5%B7%B2%E7%9F%A5%EF%BC%9Ax%5En%EF%BC%9D2%2Cy%5En%EF%BC%9D3%2C%E9%82%A3%E4%B9%88%EF%BC%88x%26sup2%3By%EF%BC%89%5E2n%EF%BC%9D________)
计算①(x²-1)(x+1)(x^4+1)(x-1)因式分解①(a+3)(a+7)+25还有一道填空已知:x^n=2,y^n=3,那么(x²y)^2n=________
计算
①(x²-1)(x+1)(x^4+1)(x-1)
因式分解
①(a+3)(a+7)+25
还有一道填空
已知:x^n=2,y^n=3,那么(x²y)^2n=________
计算①(x²-1)(x+1)(x^4+1)(x-1)因式分解①(a+3)(a+7)+25还有一道填空已知:x^n=2,y^n=3,那么(x²y)^2n=________
①(x²-1)(x+1)(x^4+1)(x-1)
题目抄错了,第一个是+号吧?
原式等于(x²+1)*(x²-1)*(x^4+1)
=(x^4-1)(X^4+1)
=x^8-1
①(a+3)(a+7)+25
=a²+10a+21+25
=a²+10a+46
没的分解了,肯定哪里符号也错了.
第三题
x=2
y=3
n=1
带入原式等于12^2=144
(x²-1)(x+1)(x^4+1)(x-1)
=(x²-1)(x²-1)(x^4+1)
=(x²-1)²(x^4+1)
=(x^4-2x²+1)(x^4+1)
=x^8-2x^6+2x^4-2x²+1
(a+3)(a+7)+25
=(a+5-2)(a+5+2)+25
=(...
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(x²-1)(x+1)(x^4+1)(x-1)
=(x²-1)(x²-1)(x^4+1)
=(x²-1)²(x^4+1)
=(x^4-2x²+1)(x^4+1)
=x^8-2x^6+2x^4-2x²+1
(a+3)(a+7)+25
=(a+5-2)(a+5+2)+25
=(a+5)²+21(实数范围内不能分解了)
=(a+5+√21i)(a+5-√21i)
x^n=2,y^n=3,
(x²y)^2n
=x^4n*y^2n
=2^4*3²
=16*9
=144
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