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1. \(x^4-2x^2+1=\left(x^2-1\right)^2\)
2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
3. \(16x^2-8x+1=\left(4x-1\right)^2\)
4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)
5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)
6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)
7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)
h, \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(\Rightarrow\left(27x^3-8\right):\left(3x-2\right)\\ =\left(3x-2\right)\left(9x^2+6x+4\right):\left(3x-2\right)\\ =9x^2+6x+4\)
g, \(x^4-2x^2+1=\left(x^2-1\right)^2\)
\(\Rightarrow\left(x^4-2x^2+1\right):\left(1-x^2\right)\\ =\left(x^2-1\right)^2:\left(1-x^2\right)\\ =x^2-1\)
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: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)
\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)
\(=-15x^3y^2+25x^2y^2-5xy^3\)
f) = x2( x - 4 ) - 9( x - 4 ) = ( x - 4 )( x - 3 )( x + 3 )
g) = 4( x - y ) + ( x - y )2 = ( x - y )( x - y + 4 )
h) = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
i) = ( x - y )( x + y ) - 4( x + y ) = ( x + y )( x - y - 4 )
j) = ( x - y )( x2 + xy + y2 ) - 3( x - y ) = ( x - y )( x2 + xy + y2 - 3 )
Trả lời:
f, x3 - 4x2 - 9x + 36 = ( x3 - 4x2 ) - ( 9x - 36 ) = x2 ( x - 4 ) - 9 ( x - 4 ) = ( x - 4 )( x2 - 9 ) = ( x - 4 )( x - 3 )( x + 3 )
g, 4x - 4y + x2 - 2xy + y2 = ( 4x - 4y ) + ( x2 - 2xy + y2 ) = 4 ( x - y ) + ( x - y )2 = ( x - y ) ( 4 + x - y )
h, x4 + x3 + x2 - 1 = ( x4 + x3 ) + ( x2 - 1 ) = x3 ( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
i, x2 - y2 - 4x - 4y = ( x2 - y2 ) - ( 4x + 4y ) = ( x - y )( x + y ) - 4 ( x + y ) = ( x + y )( x - y - 4 )
j, x3 - y3 - 3x + 3y = ( x3 - y3 ) - ( 3x - 3y ) = ( x - y )( x2 + xy + y2 ) - 3 ( x - y ) = ( x - y )( x2 + xy + y2 - 3 )
6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)
8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2 = -(y-x^2-x) (y+x^2+x)
10, 4x mũ 2 ( x + y ) -x - y = (2x-1) (2x+1) (y+x)
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
1, \(7x^2y^5-14x^3y^4-21y^3=7y^3\left(x^2y^2-2x^3y-3\right)\)
2, \(-12x^2y+6xy^2-15xy=3xy\left(-4x+2y-5\right)\)
3, \(-14x^2y-21xy^2+28x^2y^2=-7xy\left(2x+3y-4xy\right)\)
A=(12x3y4-15x2y3-9x3y2):3x2y
=(12x3y4:3x2y)-(15x2y3:3x2y)- (9x3y2:3x2y)
=4x2y3-5y2-3x2
B, sai đề bạn ơi