Khai triển các tích sau
a) (3x+5)^2
b) (x^2-4y)^2
c) (8y+1) (8y-1)
d) (2x^3+1)
e) 27y^3-8
f) 125+27y^3
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D) 64x^3-1/8y^3
= (4x)^3 + (1/2y)^3
= ( 4x + 1/2y ) [ (4x)^2 - 4x.1/2y + (1/2y)^2 ]
E) 125x^6-27y^9
( câu này mik chưa rõ nên vx chưa tek giải cho bn )
HOk tốt nhé
1/ x^2 +4xy +4y^2 = (x +2y)^2
2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3
3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3
4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
1, \(x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)
2, đề sai
3, \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)
tương tự ...
8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
Câu 2 đề ko sai nha bạn.
2) x2 - (\(\sqrt{y^3}\))2 ( y>0)
= ( x -\(\sqrt{y^3}\)) ( x +\(\sqrt{y^3}\))
\(x^3+27y^3=x^3+\left(3y\right)^3=\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)
\(a^6-8b^3=\left(a^2\right)^3-\left(2b\right)^3=\left(a^2-2b\right)\left(a^4+2a^2b+4b^2\right)\)
\(8y^3-125=\left(2y\right)^3-5^3=\left(2y-5\right)\left(4y^2+10y+25\right)\)
\(8z^3+x^3=\left(2z\right)^3+x^3=\left(2z+x\right)\left(4z^2-2xz+x^2\right)\)
1. x3 + 8 = (x + 2 )(x2 - x + 1)
2. 27 - 8y3 = ( 3 - 2y ) ( 9 + 6y + 4y2 )
3. y6 + 1 = (y2)3 + 1 = ( y2 + 1) ( y4 - y2 +1 )
4.64x3 - \(\dfrac{1}{8}\)y3 = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x2 + 2xy + \(\dfrac{1}{4}\)y2)
5. 125x6 - 27y9 = (5x2)3 - (3y3)3
= ( 5x2 - 3y3)(25x4 +15x2y3 + 9y6)
Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
Ta có : x2 + 3x
= x2 + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
\(=\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
a) ( 3x + 5 )2 = 9x2+30x+25
b) ( x2- 4y )2 = x4 - 8x2y + 16y2
c) ( 8y+1 )( 8y-1 ) = 64y2 - 1
d) ( 2x3+1 ) = 8x9+6x6+6x3+1
e) 27y3 - 8 = ( 3y )3 - 23 = ( 3y -2 )( 9y2+6y+4 )
f)125 + 27y3 = 53 + ( 3y )3 = ( 5+3y )( 25+30y+9y2 )
Hk tốt