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hoganjohnson007
@hoganjohnson007
August 2022
1
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помогите пожалуйста с 13 и 14
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sedinalana
13
1)p²/q-q²/p=(p³-q³)/pq
2)1/(p+q)*(p³-q³)/hq=(p³-q³)/pq(p+q)
3)(p³-q³)/pq-(p³-q³)/(p+q)(pq=(p³+p²q-pq²-q³-p³+q³)/(p+q)pq=
=pq(p-q)/(p+q)pq=(p-q)/(p+q)
4)(p-q)/(p+q):(p-q)/p=(p-q)/(p+q)*p/(p-q)=p/(p+q)
14
1)1/b²-1/c²=(c²-b²)/b²c²
2)(b²+c²)/b²c²*(c²-b²)/b²c²=(c^4-b^4)/b^4c^4
3)1/a²-1/c²=(c²-a²)/a²c²
4)(c²-a²)/a²c²*(a²+c²)/a²c²=(c^4-a^4)/a^4c^4
5)(c^4-b^4)/b^4c^4-(c^4-a^4)/a^4c^4=(a^4c^4-a^4b^4-b^4c^4+a^4b^4)/a^4b^4c^4=c^4(a^4-b^4)/a^4b^4c^4(a^4-b^4)/a^4b^4
6)(a^4-b^4)/a^4b^4 :(a²+b²)/a²b²=(a²-b²)(a²+b²)/a^4b^4c^4 *a²b^2/(a²+b²)=
=(a²-b²)/a²b²
2 votes
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Answers & Comments
1)p²/q-q²/p=(p³-q³)/pq
2)1/(p+q)*(p³-q³)/hq=(p³-q³)/pq(p+q)
3)(p³-q³)/pq-(p³-q³)/(p+q)(pq=(p³+p²q-pq²-q³-p³+q³)/(p+q)pq=
=pq(p-q)/(p+q)pq=(p-q)/(p+q)
4)(p-q)/(p+q):(p-q)/p=(p-q)/(p+q)*p/(p-q)=p/(p+q)
14
1)1/b²-1/c²=(c²-b²)/b²c²
2)(b²+c²)/b²c²*(c²-b²)/b²c²=(c^4-b^4)/b^4c^4
3)1/a²-1/c²=(c²-a²)/a²c²
4)(c²-a²)/a²c²*(a²+c²)/a²c²=(c^4-a^4)/a^4c^4
5)(c^4-b^4)/b^4c^4-(c^4-a^4)/a^4c^4=(a^4c^4-a^4b^4-b^4c^4+a^4b^4)/a^4b^4c^4=c^4(a^4-b^4)/a^4b^4c^4(a^4-b^4)/a^4b^4
6)(a^4-b^4)/a^4b^4 :(a²+b²)/a²b²=(a²-b²)(a²+b²)/a^4b^4c^4 *a²b^2/(a²+b²)=
=(a²-b²)/a²b²