khai triển đẳng thức
9y^2-(x+y)^2
có công thức
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= (x+1-y)(x+1+y)
hằng đẳng thức số 3: a^2 - b^2 = (a-b)(a+b)
\(\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(\hept{\begin{cases}\frac{x+3}{x^2+4x+4}\\\frac{x+1}{3x+6}\end{cases}}MTC:3\left(x+2\right)^2\)
\(\Rightarrow\hept{\begin{cases}\frac{x+3}{x^2+4x+4}=\frac{3\left(x+3\right)}{3\left(x+2\right)^2}=\frac{3x+9}{3\left(x+2\right)^2}\\\frac{x+1}{3x+6}=\frac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)^2}=\frac{x^2+3x+2}{3\left(x+2\right)^2}\end{cases}}\)
a) (x+y)3 - x3 - y3
= x3 + 3x2y + 3xy2 + y3 - x3 -y3
= 3x2y+ 3xy2
= 3xy(x+y)
b) 4x2 - y2 + 4x + 1
= (2x)2 + 2. 2x.1 + 12 - y2
= (2x+1)2 - y2
= (2x+1-y)(2x+1+y)
a) (1-5a)^2 = 1- 10a + 25a^2
b) x^2 - 4 = (x-2)(x+2)
c) 1-x^2 = (1-x)(1+x)
d) 4a^2 - 9= (2a - 3)(2a + 3)
\(\frac{3x}{2x+4}=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}=\frac{3x^2-6x}{2\left(x+2\right)\left(x-2\right)}=\)
\(\frac{x+3}{x^2-4}=\frac{2\left(x+3\right)}{2\left(x+2\right)\left(x-2\right)}=\frac{2x+6}{2\left(x+2\right)\left(x-2\right)}\)
`(3x)/(2x+4)`
`= (3x (x-2) )/( (2x+4) (x-2) )`
`= (3x^2 - 6x)/(2 (x+2) (x-2) )`
`(x+3)/(x^2 - 4)`
`= (x+3)/(x^2 - 2^2)`
`= (x+3)/( (x-2) (x+2) )`
`= (2 (x+3) )/(2 (x-2) (x+2) )`
`= (2x+ 6)/(2 (x+2) (x-2) )`
\(\left(x-\frac{1}{2}\right)^2=x^2-2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2\)
\(=x^2-x+\frac{1}{4}\)
\(9y^2-\left(x+y\right)^2=\left(3y-x-y\right)\left(3y+x+y\right)=\left(2y-x\right)\left(4y+x\right)=8y+2xy-4xy-x^2\)
\(=8y^2-2xy-x^2\)
`9y^2 - (x+y)^2`
`= (3y)^2 - (x+y)^2`
`= (3y - x - y) (3y +x+y)`
`= (2y - x) (4y+x)`
`= 2y (4y+x)-x(4y+x)`
`= 8y^2 + 2xy - 4xy - x^2`
`= 8y^2 - 2xy - x^2`
Vận dụng HĐT : `A^2 - B^2 = (A-B)(A+B)`