viết biểu thức sau dưới dạng tổng của 2 bình phương: x^2 + 4y^2 -6x + 4y +10
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a) x2 + 4y + 4y2 + 26 - 10x = ( x2 - 10x + 25 ) + ( 4y2 + 4y + 1 ) = ( x - 5 )2 + ( 2y + 1 )2
b) 4y2 + 34 - 10x + 12y + x2 = ( x2 - 10x + 25 ) + ( 4y2 + 12y + 9 ) = ( x - 5 )2 + ( 2y + 3 )2
c) -10x + y2 - 8y + x2 + 41 = ( x2 - 10x + 25 ) + ( y2 - 8y + 16 ) = ( x - 5 )2 + ( y - 4 )2
d) x2 + 9y2 - 12y + 29 - 10x = ( x2 - 10x + 25 ) + ( 9y2 - 12y + 4 ) = ( x - 5 )2 + ( 3y - 2 )2
a) \(x^2+4y+4y^2+26-10x\)
\(=\left(x^2-10x+25\right)+\left(4y^2+4y+1\right)\)
\(=\left(x-5\right)^2+\left(2y+1\right)^2\)
b) \(4y^2+34-10x+12y+x^2\) đề ntn à?
\(=\left(4y^2+12y+9\right)+\left(x^2-10x+25\right)\)
\(=\left(2y-3\right)^2+\left(x-5\right)^2\)
c) \(-10x+y^2-8y+x^2+41\)
\(=\left(x^2-10x+25\right)+\left(y^2-8y+16\right)\)
\(=\left(x-5\right)^2+\left(y-4\right)^2\)
d) \(x^2+9y^2-12y+29-10x\)
\(=\left(x^2-10x+25\right)+\left(9y^2-12y+4\right)\)
\(=\left(x-5\right)^2+\left(3y-2\right)^2\)
a) \(2x^2+2b^2=x^2+b^2+x^2+b^2=x^2+2xb+b^2+x^2-2xb+b^2=\left(x+b\right)^2+\left(x-b\right)^2\)
a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(5x+\frac{1}{2}y\right)^2\)
b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)
c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)
\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)
d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
\(\left(2x-4y\right)^2+2\left(2x-4y\right)+1=\left(2x-4y+1\right)^2\)
x^2 + 4y^2 -6x + 4y +10
=x2-6x+9+4y2+4y+1
=(x-3)2+(2y+1)2
x^2-4y^2-6x+4y+10
= x^2- 2.x.3 + 9 + 4y^2-2.2y+1
= ( x - 3)^2+ (2y-1) ^2