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\(a,\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)
\(=a^2+4x+4-a^2+4\)
\(=4x+8\)
\(=4\left(x+2\right)\)
\(b,\left(a+b\right)^2-\left(a-b\right)^2\)
\(=a^2+2ab+b^2-\left(a^2-2ab+b^2\right)\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=4ab\)
\(c,\left(3x+4\right)^2-10x-\left(x+4\right)\left(x-4\right)\)
\(=9x^2+24x+16-10x-x^2+16\)
\(=8x^2+14x+32\)
\(=2\left(4x^2+7x+16\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P = 9x2 + 24x + 16 -10x - x2 +16
P = 8x2 +14x +32
P = 2(4x2 + 7x +16)
P = ( 3x + 4 )2 - 10x - ( x - 4 )( x + 4 )
P = 9x2 + 24x + 16 - 10x - ( x2 - 16 )
P = 9x2 + 24x + 16 - 10x - x2 + 16
P = 8x2 + 14x + 32
P = 2( 4x2 + 7x + 16 )
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x+2)2+x(x-4)
=x2+4x+4+x2-4x
=2x2+4
b)(x-3)2-(x+3)(x-4)
=x2-6x+9-x2+4x-3x+12
=-5x+12
c) (3x+1)2+3x(2-4x)
=9x2+6x+1+6x-12x2
=-3x2+12x+1
d) (2x-4y)2-(2x-3)(2x-3y)
=4x2-16xy+16y2-4x2+6xy+6x-9y
=16y2-10xy+6x-9y
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)
\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)
\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)
\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)
\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)
b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)
Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)
2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)
3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
D = (3x - 2)^2 - 3(x - 4)(4 + x) + (x - 3)^3 - (x^2 - x + 1)(x + 1)
D = 9x^2 - 12x + 4 - 3x^2 + 48 + x^3 - 9x^2 + 27x - 27 - x^3 - 1
D = -3x^2 + 15x + 24
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
1.
$M=(x^2+6x+9)+(x^2-9)-2(x^2-2x-8)$
$=x^2+6x+9+x^2-9-2x^2+4x+16=(x^2+x^2-2x^2)+(6x+4x)+(9-9+16)$
$=10x+16=5(2x+1)+11=5.0+11=11$
2.
$V=(9x^2+24x+16)-(x^2-16)-10x=9x^2+24x+16-x^2+16-10x$
$=(9x^2-x^2)+(24x-10x)+(16+16)=8x^2+14x+32$
$=8(\frac{-1}{10})^2+14.\frac{-1}{10}+32=\frac{767}{25}$
3.
$P=(x^2+2x+1)-(4x^2-4x+1)+3(x^2-4)$
$=x^2+2x+1-4x^2+4x-1+3x^2-12$
$=(x^2-4x^2+3x^2)+(2x+4x)+(1-1-12)$
$=6x-12=6.1-12=-6$
4.
$Q=(x^2-9)+(x^2-4x+4)-2x^2+8x$
$=x^2-9+x^2-4x+4-2x^2+8x$
$=(x^2+x^2-2x^2)+(-4x+8x)-9+4$
$=4x-5=4(-1)-5=-9$
\(\left(3x+4\right)^2-10x-\left(x-4\right)\left(x+4\right)\)
\(=9x^2+24x+16-10x-x^2+16\)
\(=8x^2+14x\)
P = ( 3x + 4 )2 - 10x - ( x - 4 )( x + 4 )
P = 9x2 + 24x + 16 - 10x - ( x2 - 16 )
P = 9x2 + 24x + 16 - 10x - x2 + 16
P = 8x2 + 14x + 32
P = 2( 4x2 + 7x + 16 )