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a: \(=25x^4-10x^3+5x^2\)
c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
a) 2( x - 1 )2 - 4( 3 + x )2 + 2x( x - 5 )
= 2( x2 - 2x + 1 ) - 4( 9 + 6x + x2 ) + 2x2 - 10x
= 2x2 - 4x + 2 - 36 - 24x - 4x2 + 2x2 - 10x
= ( 2x2 - 4x2 + 2x2 ) + ( -4x - 24x - 10x ) + ( 2 - 36 )
= -38x - 34
b) 2( 2x + 5 )2 - 3( 4x + 1 )( 1 - 4x )
= 2( 4x2 + 20x + 25 ) + 3( 4x + 1 )( 4x - 1 )
= 8x2 + 40x + 50 + 3( 16x2 - 1 )
= 8x2 + 40x + 50 + 48x2 - 3
= 56x2 + 40x + 47
c) ( x - 1 )3 - x( x - 3 )2 + 1
= x3 - 3x2 + 3x - 1 - x( x2 - 6x + 9 ) + 1
= x3 - 3x2 + 3x - x3 + 6x2 - 9x
= 3x2 - 6x
d) ( x + 2 )3 - x2( x + 6 )
= x3 + 6x2 + 12x + 8 - x3 - 6x2
= 12x + 8
e) ( x - 2 )( x + 2 ) - ( x + 1 )3 - 2x( x - 1 )2
= x2 - 4 - ( x3 + 3x2 + 3x + 1 ) - 2x( x2 - 2x + 1 )
= x2 - 4 - x3 - 3x2 - 3x - 1 - 2x3 + 4x2 - 2x
= -3x3 + 2x2 - 5x - 5
f) ( a + b - c )2 - ( b - c )2 - 2a( b - c )
= [ ( a + b ) - c ]2 - ( b2 - 2bc + c2 ) - 2ab + 2ac
= [ ( a + b )2 - 2( a + b )c + c2 ] - b2 + 2bc - c2 - 2ab + 2ac
= a2 + 2ab + b2 - 2ac - 2bc + c2 - b2 + 2bc - c2 - 2ab + 2ac
= a2
a) \(2\left(x-1\right)^2-4\left(3+x\right)^2+2x\left(x-5\right)\)
Dùng hẳng đẳng thức thứ nhất + hai :
= \(2\left(x^2-2\cdot x\cdot1+1^2\right)-4\left(3^2+2\cdot3\cdot x+x^2\right)+2x^2-10x\)
= \(2\left(x^2-2x+1\right)-4\left(9+6x+x^2\right)+2x^2-10x\)
= \(2x^2-4x+2-36-24x-4x^2+2x^2-10x\)
= \(-38x-34\)
b) 2(2x + 5)2 - 3(4x + 1)(1 - 4x)
Dùng đẳng thức thứ 1 + 3
= 2[(2x)2 + 2.2x.5 + 52 ] - (-3)[(4x)2 - 12 ]
= 2(4x2 + 20x + 25) - (-3).(16x2 - 1)
= 8x2 + 40x + 50 - (3 - 48x2)
= 8x2 + 40x + 50 - 3 + 48x2
= 56x2 + 40x + 47
c) (x - 1)3 - x(x - 3)2 + 1
Dùng đẳng thức 2 + 5:
= x3 - 3.x2.1 + 3.x.12 - 13 - x(x2 - 2.x.3 + 32) + 1
= x3 - 3x2 + 3x - 1 - x3 + 6x2 - 9x + 1
= (x3 - x3) + (-3x2 + 6x2) + (3x - 9x) + (-1 + 1)
= 3x2 - 6x
d) (x + 2)3 - x2(x + 6)
= x3 + 3.x2.2 + 3.x.22 + 23 - x3 - 6x2
= x3 + 6x2 + 12x + 8 - x3 - 6x2
= (x3 - x3) + (6x2 - 6x2) + 12x + 8 = 12x + 8
e) Dùng đẳng thức thứ 3,4 và 2
= x2 - 4 - (x3 + 3.x2.1 + 3.x.12 + 13) - 2x(x2 - 2.x.1 + 12)
= x2 - 4 - (x3 + 3x2 + 3x + 1) - 2x3 + 4x2 - 2x
= x2 - 4 - x3 - 3x2 - 3x - 1 - 2x3 + 4x2 - 2x
= (x2 - 3x2 + 4x2) + (-4 - 1) + (-x3 - 2x3) + (-3x - 2x)
= 2x2 - 5 - 3x3 - 5x
f) Đặt \(a+b-c=A\)
\(b-c=B\)
= \(A^2-B^2-2AB\)
= \(A^2-2AB+\left(-B\right)^2\)
\(=A^2-2AB+B^2\)
= (A - B)2
= (a + b - c - (b - c))2
= (a + b - c - b + c)2
= a2
Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
Bài 1:
a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)
\(=\dfrac{15x^2y^2z}{3xyz}\)
\(=5xy\)
b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)
\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)
\(=15x^4-12x^3+9x^2\)
c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)
\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)
\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)
\(=2x+5+\dfrac{20}{x-4}\)
d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)
\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)
\(=-15x^3y^2+25x^2y^2-5xy^3\)
a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)
b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)
\(=\left(3x-5\right)\left(x+1\right)\)
a: \(-2y\left(5xy+3x^2\right)=-10xy^2-6x^2y\)
b: \(\left(2x+y\right)\left(4x-y\right)\)
\(=8x^2-2xy+4xy-y^2\)
\(=8x^2+2xy-y^2\)