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a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
\(4x^2-\frac{1}{9}\left(y+1\right)^2=\left(2x\right)^2-\left(\frac{1}{3}\left(y+1\right)\right)^2\)
\(=\left(2x-\frac{1}{3}\left(y+1\right)\right)\left(2x+\frac{1}{3}\left(y+1\right)\right)\)
\(=\left(2x-\frac{1}{3}y-\frac{1}{3}\right)\left(2x+\frac{1}{3}y+\frac{1}{3}\right)\)
a/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
\(a.=\left(2x\right)^2-2.2x.2y+\left(2y\right)^2=\left(2x-2y\right)^2\)
\(b.=\left(3x\right)^2-2.3x.2+2^2=\left(3x-2\right)^2\)
a. 4x2+4y2-8xy=(2x)2+(2y)2-8xy
=(2x-2y)2
b.9x2-12x+4=(3x)2-12x+22
=(3x-2)2
c.xy2+1/4x2y4+1=xy2+(1/2xy2)2+1
=(1/2xy2+2)2
Bài làm:
Ta có: \(\frac{9}{4x^2}+\frac{9y^2}{4}-\frac{9y}{2x}\)
\(=\left(\frac{3}{2x}\right)^2-2.\frac{3}{2x}.\frac{3y}{2}+\left(\frac{3y}{2}\right)^2\)
\(=\left(\frac{3}{2x}-\frac{3y}{2}\right)^2\)
Áp dụng bất đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\) với a = 2x + 3y , b = 1
Được : \(\left(2x+3y\right)^2+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
\(1.z^2-6z+5-t^2-4t\)
\(=\left(z^2-6z+9\right)-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
\(3,x^2-2xy+2y^2+2y+1\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
\(4x^2+16x+7-y^2-6y=\left(4x^2+16x+16\right)-\left(y^2+6y+9\right)=\left(2x+4\right)^2-\left(y+3\right)^2\)
\(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)