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1. 20\(x^2y^3\) : 4x\(y^2\) = 5xy
2. \(\dfrac{-1}{2}x^4y^4\) : \(\dfrac{2}{3}x^2y^2\) = \(\dfrac{-3}{4}x^2y^2\)
3. \(\left(-xy\right)^6:\left(-xy\right)^2=\left(-xy\right)^2\) = xy
4. 27\(x^2y^3z^4:\left(-3xyz\right)^2\) = 27\(x^2y^3z^4\) : 9 \(x^2y^2z^2\) = 3y\(z^2\)
5. \(\left(-x\right)^{10}:\left(-x\right)^5=\left(-x\right)^2\) = x
a: \(=\dfrac{5x^2y^4}{-10x^2y}=-\dfrac{1}{2}y^3=-\dfrac{1}{2}\cdot8=-4\)
b: \(=\dfrac{15x^4y^2}{5x^3y}+\dfrac{20x^3y^2}{5x^2y}=3xy+4xy=7xy\)
\(=7\cdot\dfrac{1}{7}\cdot2009=2009\)
1. \(\left(-a\right)^7\) : \(a^5\) = \(\left(-a\right)^2\) = a
2. 28 \(y^4z^3\) : 14 \(y^3z^2\) = 2yz
3. 25\(a^2bc^2\) : 5abc = 5ac
1: \(\Leftrightarrow3x+4x=4\)
=>7x=4
hay x=4/7
2: \(\Leftrightarrow3x-5x-5^3:5^2=0\)
=>-2x=5
=>x=-5/2
a) 25x² - 10xy + y²
= (5x)² - 2.5x.y + y²
= (5x - y)²
b) 4/9 x² + 20/3 xy + + 25y²
= (2/3 x)² + 2.2/3 x.5y + (5y)²
= (2/3 x + 5y)²
c) 9x² - 12x + 4
= (3x)² - 2.3x.2 + 2²
= (3x - 2)²
d) Sửa đề: 16u²v⁴ - 8uv² + 1
= (4uv²)² - 2.4uv².1 + 1²
= (4uv² - 1)²
\(25x^2-10xy+y^2=\left(5x\right)^2-2.5x.y+y^2=\left(5x-y\right)^2\)
\(\dfrac{4}{9}x^2+\dfrac{20}{3}xy+25y^2=\left(\dfrac{2}{3}x\right)^2+2.\dfrac{2}{3}x.5y+\left(5y\right)^2=\left(\dfrac{2}{3}x+5y\right)^2\)
1: \(=\dfrac{x^2-1}{x\left(x^2-1\right)}=\dfrac{1}{x}\)
2: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{y\left(x-2\right)}=\dfrac{x+2}{y}\)
3: \(=\dfrac{2x^2+2xy-xy-y^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)\left(2x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x-y}{x-y}\)
4: \(=\dfrac{x\left(x^2-1\right)}{x\left(x^2-x-2\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{x-1}{x-2}\)
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