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1) \(x^2\left(x-1\right)+16\left(1-x\right)\)
\(=x^2\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x^2-16\right)\left(x-1\right)\)
\(=\left(x^2-4^2\right)\left(x-1\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)
2)\(81x^2-6yz-9y^2-z^2\)
\(=81x^2-9y^2-6yz-z^2=81x^2-\left(9y^2+6yz+z^2\right)=81x^2-\left(\left(3y\right)^2+6yz+z^2\right)\)
\(=81x^2-\left(3y+z\right)^2=\left(9x\right)^2-\left(3y+z\right)^2\)
\(=\left(9x+3y+z\right)\left(9x-\left(3y+z\right)\right)=\left(9x+3y+z\right)\left(9x-3y-z\right)\)
3)\(xz-yz-x^2+2xy+y^2\)
\(=z\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=z\left(x-y\right)-\left(x-y\right)^2=\left(z+x-y\right)\left(x-y\right)\)
\(a,\left(x+4\right).\left(x^2-4x+16\right)=x^3-4x^2+16x+4x^2-16x+64\) \(64\)
\(=x^3+64\)
hoặc \(\left(x+4\right).\left(x^2-4x+16\right)=x^3+64\) ÁP Dụng hằng đẳng thức
\(b,\left(x-3y\right).\left(x^2+3xy+9y^2\right)=x^3-27\)
a) (x+4).(x^2-4x+16)
= (x+4).(x^2-x.4+4^2)
= x^3+4^3
= x^3+64
b) (x-3y).(x^2+3xy+9y^2)
= (x-3y).(x^2+x.3y+(3y)^2)
= x^3-(3y)^3
= x^3-27y^3
a) Ta có x2 + 9y2 - 6xy = (x - 3y)2 (1)
Thay x = 16 ; y = 2 vào (1) ta có
(x - 3y)2 = (16 - 2.3)2 = 102 = 100
b) Ta có x3 - 6x2y + 12xy2 - 8y3
= (x - 2y)3 (1)
Thay x = 14 ; y = 2 vào (1) ta có
(x - 2y)3 = (14 - 2.2)3 = 103 = 1000
a) \(x^2+9y^2-6xy=\left(x-3y\right)^2\)
Thay \(x=16;y=2\)vào biểu thức trên ta có :
\(\left(16-3.2\right)^2=\left(16-6\right)^2=10^2=100\)
Vậy tại x = 16 và y = 2 thì biểu thức trên = 100
b) \(x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
Thay x = 14 và y = 2 vào biểu thức trên ta có :
\(\left(14-2.2\right)^3=\left(14-4\right)^3=10^3=1000\)
Vậy tại x = 14 và y = 2 thì biểu thức trên = 1000
16) 2x + 2y - x2 - xy = ( 2x + 2y ) - ( x2 + xy ) = 2( x + y ) - x( x + y ) = ( x + y )( 2 - x )
17) x2 - 2x - 4y2 - 4y = ( x2 - 4y2 ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2( x + 2y ) = ( x + 2y )( x - 2y - 2 )
18) x2y - x3 - 9y + 9x = ( x2y - x3 ) - ( 9y - 9x ) = x2( y - x ) - 9( y - x ) = ( y - x )( x2 - 9 ) = ( y - x )( x - 3 )( x + 3 )
19) x2( x - 1 ) + 16( 1 - x ) = x2( x - 1 ) - 16( x - 1 ) = ( x - 1 )( x2 - 16 ) = ( x - 1 )( x - 4 )( x + 4 )
20) 2x2 + 3x - 2xy - 3y = ( 2x2 - 2xy ) + ( 3x - 3y ) = 2x( x - y ) + 3( x - y ) = ( x - y )( 2x + 3 )
20, \(2x^2+3x-2xy-3y=2x\left(x-y\right)+3\left(x-y\right)=\left(2x+3\right)\left(x-y\right)\)
16, \(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
17, \(x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x-2y-2\right)\left(x+2y\right)\)
18, \(x^2y-x^3-9y+9x=-x\left(x^2-9\right)+y\left(x^2-9\right)=\left(-x-y\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
19, \(x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)
\(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}xy\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
Giải:
+) \(\left(x-4\right)\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4.x+4^2\right)\)
\(=x^3-4^3\)
\(=x^3-64\)
+) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
+) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
+) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
Vậy ...
1) ( x - 2 ) ( x2 + 2x +4 )=\(x^3-8\)
2) ( x+4) ( x2 - 4x + 16 ) \(=x^3+64\)
3) ( x - 3y ) (x2 + 3xy + 9y2 ) \(=x^3-27y^3\)
4) ( x2 - 1/3 ) ( x4 + 1/3x2 + 1/9 ) \(=x^6-\frac{1}{27}\)
x2-x3-9y+9x=-x3+x2+9x-9-9y+9=(x-1)(-x2+9-9)=-x2(x-1).
x2(x-1)+16(x-1)=(x-1)(x2+16)