Question

已知x+y=6,xy=7,求值:①x^2+y^2②x^2-xy+y^2,③x^2 y+xy^2,

④x-y.(过程)>∽< 三克油(Thank you!!!)

Answer 1

①x^2+y^2=(x+y)²-2xy=6²-2*7=36-14=22;
②x^2-xy+y^2=22-7=15;
③x^2 y+xy^2=xy(x+y)=7*6=42;
④x-y.
(x-y)²=x²+y²-2xy=22-2*7=8
x-y=±√8=±2√2；

Answer 2

解
x^2+y^2=（x+y）^2-2xy=36-14=22
x^2-xy+y^2=（x+y）^2-3xy=36-21=15
x^2y+xy^2=xy（x+y）=6*4=42
（x+y）^2-4xy=（x-y）^2=36-28=8
所以（x-y）^2=8，x-y=±2√2

Answer 3

已知x+y=6,xy=7

①x²+y²
=x²+2xy+y²-2xy
=﹙x+y﹚²-2xy
=6²-14
=22

②x²-xy+y²
=x²+2xy+y²-3xy
=﹙x+y﹚²-3xy
=6²-21
=15

③x²y+xy²
=xy（x+y）
=7×6
=42

④(x-y)²
=x²+y²-2xy
=22-2×7=8

∴x-y=±√8=±2√2

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