rút gọn biểu thức
2 . ( x - y ) . ( x + y ) + ( x + y ) ^ 2 + ( x - y ) ^ 2
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Lời giải:
$x(x+y)-y(x+y)+x^2+y^2=(x-y)(x+y)+x^2+y^2$
$=x^2-y^2+x^2+y^2=2x^2$
\(A=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4\left(y^2-1\right)\)
\(=\left(x-y-x-y\right)^2-4\left(y^2-1\right)\)
\(=\left(-2y\right)^2-4y^2+4=4\)
(x + y + z)2 – 2.(x + y + z).(x + y) + (x + y)2
= [(x + y + z) – (x + y)]2 (Áp dụng HĐT (2) với A = x + y + z ; B = x + y)
= z2.
\(\left(a\right):\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\\ =x^2+2xy+y^2-x^2+2xy-y^2\\ =4xy\)
\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)
\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)
\(a,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=4xy\\ b,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)=\left(x+y-x+y\right)^2=4y^2\\ c,\left(x^2-1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x^3+1\right)\\ =x^4-x^3+x-1\)
a. (x + y)2 - (x - y)2
= (x + y - x + y)(x + y + x - y)
= 2y . 2x
= 4xy
b. (x + y)2 + (x - y)2 - 2(x + y)(x - y)
= (x2 + 2xy + y2) + (x2 - 2xy + y2) - 2(x2 - y2)
= x2 + 2xy + y2 + x2 - 2xy + y2 - 2x2 + 2y2
= x2 + x2 - 2x2 + 2xy - 2xy + y2 + y2 + 2y2
= 4y2
c. (x2 - 1)(x2 - x + 1)
= x4 - x3 + x2 - x2 + x - 1
= x4 - x3 + x - 1
\(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)
\(D=\dfrac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\dfrac{y^2\left(x+y\right)}{x}\\ D=\dfrac{\left(x^2-y^2\right)\left(x+y\right)+y^2\left(x+y\right)}{x}\\ D=\dfrac{\left(x+y\right)\left(x^2-y^2+y^2\right)}{x}=\dfrac{x^2\left(x+y\right)}{x}=x\left(x+y\right)\)
A = x ( x + y ) - y ( x + y )
A = ( x + y ) ( x - y )
A = x\(^2\) - y\(^2\)
Tại x = \(\dfrac{-1}{2}\) và y = -2 ta có
\(\left(\dfrac{-1}{2}\right)^2-\left(-2\right)^2\) \(=\) \(\dfrac{-15}{4}\)
= 2(x^2-y^2) + x^2 + 2xy + y^2+x^2-2xy+y^2
= 2x^2 - 2y^2 + x^2 + 2xy + y^2 + x^2 - 2xy + y^2
= 4x^2
Theo mình là :
2 ( x-y )(x+y)+(x+y)2+(x-y)2 = (2x-2y) (x+y) + (x+y)(x+y) + (x-y)(x-y)
= (x-y)(x+y) + x2+y2 + x2 - 2xy + y2
= x2 - y2 + x2 +y2 + (x-y)2