答：ax+by=7，两边乘以x，得：ax^2+byx=7x 与ax^2+by^2=49两边相减可得：by(x-y)=7x-49……（1）同理可得：by^2(x-y)=49x-133……（2）同理可得：by^3(x-y)=133x-406……（3）由（1）和（2）可得：y=(49x-133)/(7x-49)……（4）由（2）和（3）可得：y=（133x-406...
(2)首先将原式重新分组进行因式分解,进而代入ay+bx,ax+by求出即可. 解答: 解:(1)∵a+b=x+y=3, ∴(a+b)(x+y)=9, ∴(ax+by)+(ay+bx)=9, ∵ax+by=7, ∴ay+bx=2; (2)∵ax+by=7,ay+bx=2, ∴(a 2 +b 2)xy+ab(x 2 +y 2) =xya 2 +xyb 2 +abx 2 +aby...
得3a-4b=7①,把x=1,y=2代入ax-by=1中,得a-2b=1②,解由①②组成的方程组得, a=5 b=2 . 由方程组的定义,可知甲的解答 x=3 y=4 满足原方程,代入后,可得a,b间的一个关系式3a-4b=7,乙求出的解不满足原方程,而满足方程ax-by=1,代入后可得a,b的另一个关系式a-2b=1,从而可求出a,b的...
∵ S_3=ax^3+by^3= ( (x+y) ) ( (ax^2+by^2) )-xy ( (ax+by) )= ( (x+y) )S_2-xyS_1,S_4=ax^4+by^4= ( (x+y) ) ( (ax^3+by^3) )-xy ( (ax^2+by^2) )= ( (x+y) )S_3-xyS_2, ∴ \( (((array)(ll) (49 ( (x+y) )-7xy=133) \ (13...
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得49(x+y)=133+7xy133(x+y)=(ax³+by³)(x+y)=ax^4+(ax²+by²)xy+by^4=406+149xy得133(x+y)=406+49xy由以上三个方程解得x+y=5/2,xy=-3/2,a+b=21得所求=4832 解析看不懂?免费查看同类题视频解析查看解答 相似问题 已知ax+by=7,ax2+by2=49,ax3+by3=133,ax4+by4=...
(x+y) 由 ax^2+by^2=49 ,两边同乘以(x+y),即(ax^2+by^2)(x+y)=49(x+y) 整理,得ax^3+by^3+xy(ax+by)=49(x+y) ∵ax^3+by^3=133,ax+by=7 ∴-133+7xy=49(x+y) ②用同样的方法,把 ax^3+by^3=133 两边同乘以(x+y),整理后,得:406+49xy=133(x+y)③由②,...
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【解析】 由题意得: 2a+b=-2 t 2a-6b=7 1 解得: a= 2 b=-1 所以原方程组为: 2 -y=7 -x+y=-2 10 = 解得: 3 16 y 3 故答案为: 10 = 3 a=- 2 ,b=-1;方程组的解为: 16 y 3 : 过程略. 结果一 题目 题目】 甲乙两位同学在解方程组X+by=7时,甲 看错了第一个方程...
(2)首先将原式重新分组进行因式分解,进而代入ay+bx,ax+by求出即可. 解答:解:(1)∵a+b=x+y=3,∴(a+b)(x+y)=9,∴(ax+by)+(ay+bx)=9,∵ax+by=7,∴ay+bx=2;(2)∵ax+by=7,ay+bx=2,∴(a2+b2)xy+ab(x2+y2)=xya2+xyb2+abx2+aby2=ax(ay+bx)+by(bx+ay)=(ay+bx)(ax+by...