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\(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)
TL:
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)\left(2x+1-x+1\right)\)
\(=3x.\left(x+2\right)\)
a,x2-4xy+4y2
=(x-2y2
b,4x4+9y2-12x2y
=(2x2)2+(3y)2-12x2y
(2x2-3y)
\(6x-9-x^2\)
\(=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
\(=-1.\left(x-3\right)^2\)
b ) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=2x.2\left(2x+1\right)\)
\(=4x\left(2x+1\right)\)
Sao chẳng ai T z
a) 9 -(x-y)2
= 32 - (x-y)2
= (3-x+y).(3+x-y)
b) (x2 +4)2 - 16x2
= (x2+4)2 - (4x)2
= (x2 + 4 -4x).(x2 + 4 +4x)
\(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
\(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
Cái này thì bn sử dụng hằng đẳng thức là đc bạn nhé!
\(x^2+4xy+4y^2-2x-4y+1\)
\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+1\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+1\)
\(=\left(x+2y-1\right)^2\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
1
a) x2 + 4y2 + 4xy - 16
=(x2 + 4xy + 4y2) - 16
=(x+2y)2 - 16
=(x+2y-4)(x+2y+4)
b)x2 + y2 - 2x + 4y + 5 =0
<=> x2 - 2x + 1 + y2 - 4y + 4=0
<=> (x-1)2 + (y-2)2 =0
<=> x=1 và y=2
=(x^2-4x+4)-4y^2
=(x-2)^2-(2y)^2
=(x-2+2y)x(x-2-2y)
nếu thấy đúng thì k nhe
mk k lại cho
\(x^2+4xy+4y^2=\left(x+2y\right)^2\)(Nhớ k cho mình với nhé!)
\(x^2+4xy+4y^2=\left(x+2y\right)^2\)