已知(x+y)^2=6,(x-y)^2=3,求xy,x^2y^2,x^2+y^2的值
两式相减得到xy=3\/4,平方后得到x^2y^2=9\/16,将xy值代入两式中的任何一个得到x^2+y^2=9\/2
(x+y)^2=4 ,(x-y)^2=3求1.x^2y^2 2.x^2+y^2
第一题的第二小题的答案是:7\/2 过程:由(x+y)^2=x^2+2xy+y^2 =4 (1)(x-y)^2=x^2-2xy+y^2=3 (2)联立(1)和(2)可得x^2+y^2=7\/2
求值:已知(x+y)^2=324,(x-y)^2=16,求xy及x^2+y^2的值?
xy=77 [1]+[2]:2(x^2+y^2)=340 x^2+y^2=170,10,(x+y)^2-(x-y)^2 =[(x+y)+(x-y)][(x+y)-(x-y)]=2x*2y =4xy=324-16=308 所以xy=77 (x+y)^2=x^2+y^2+2xy==324 x^2+y^2=324-2xy=324-77*2=170,2,xy=77 x^2+y^2=170,2,
已知(x+y)^2=8,(x-y)^2=5,求3xy与x^2+y^2的值
(x+y)^2=8 (1)(x-y)^2=5 (2)==>(1)-(2)==>4xy=3;所以3xy=9\/4 (1)+(2)==>2x^2+2y^2=13;所以x^2+y^2=13\/2
...=3,xy+y^2=6,求①x^2+3xy+2y^2的值②x^2-y^2的值
已知x^2+xy=3,xy+y^2=6,①x^2+3xy+2y^2的值 =x²+xy+2(xy+y²)=3+2×6 =3+12 =15;②x^2-y^2的值 =(x²+xy)-(xy+y²)=3-6 =-3;请好评 如果你认可我的回答,敬请及时采纳,~如果你认可我的回答,请及时点击【采纳为满意回答】按钮~~手机提问...
已知(x+y)^2=9,(x-y)^2=5求x^2+y^2和xy值
∵(x+y)²=9 ∴x²+2xy+y²=9………① ∵(x-y)²=5 ∴x²-2xy+y²=5………② ①+②得 2(x²+y²)=14 x²+y²=7 ①-②得 4xy=4 xy=1 很高兴为您解答,祝你学习进步!【数学之美】团队为您答题。有不明白的可以追问...
初二上册数学完全平方公式1.已知(x+y^2)=16,(x-y)^2=4,求x^2+xy+y^2
x^2+xy+y^2=13 因为(x+y^2)=x^2+2XY+Y^2=16 (x-y)^2=x^2-2XY+Y^2=4 x^2+2XY+Y^2-(x^2-2XY+Y^2)=4XY=12 XY=3 x^2+2XY+Y^2+(x^2-2XY+Y^2)=2x^2+2y^2=20 又∵2(x^2+y^2)=20,∴x^2+y^2=10 ∴x^2+y^2+xy=13 ...
已知(x+y)^2=25,(x-y)^2=9,求xy与x^2+y^2的值
已知(x+y)^2=25,(x-y)^2=9,xy=((x+y)^2-(x-y)^2)\/4=4;x²+y²=25-2*4=17
已知(x+y)^2=7,(x-y)^2=9,求x^2+y^2及xy的值
(x y)^2=7,(x-y)^2=9各自展开得:x²+y²+2xy=7……… ① x²+y²-2xy=9………② ①式和②式相加可得:2(x²+y²)=16 x²+y²=8 ∴xy=-0.5
x+y=4,xy=3求(x-y)^2,x^2y+xy^2的值
回答:x+y=4,xy=3 所以(x-y)^2=(x+y)^2-4xy=16-12=4 x^2y+xy^2=xy(x+y)=3*4=12 如果不懂,请追问,祝学习愉快!