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a) \(73^2-27^2=\left(73+27\right)\left(73-27\right)=100.46=4600\)
b) \(55^2+20^2-25^2+40.45=\left(55^2-25^2\right)+\left(20^2+40.45\right)\)
\(=\left(55-25\right)\left(55+25\right)+\left(40.10+40.45\right)=30.80+40.55\)
\(=40\left(60+55\right)=40.115=4600\)
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a)\(9x^2y^2+15x^2y-21xy^2\)
=\(3xy\left(3xy+5x-7y\right)\)
b)\(3x^2\left(x+1\right)-2\left(x+1\right)\)
=\(\left(3x^2-2\right)\left(x+1\right)\)
c)\(4x^2\left(x-2y\right)-20x\left(2y-x\right)\)
=\(4x^2\left(x-2y\right)+20x\left(x-2y\right)\)
=\(\left(4x^2+20x\right)\left(x-2y\right)\)
=\(4x\left(x+5\right)\left(x-2y\right)\)
a. ta có :
\(9x^2y^2+15x^2y-21xy^2=3xy\left[3xy+5x-7y\right]\)
b.\(3x^2\left(x+1\right)-2\left(x+1\right)=\left(x+1\right)\left[3x^2-2\right]\)
\(4x^2\left(x-2y\right)+20x\left(x-2y\right)=\left(x-2y\right)\left(4x^2+20x\right)=4x\left(x-2y\right)\left(x+5\right)\)
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Trả lời:
a) x2 + 4y2 + 4xy = x2 + 2.x.2y + (2y)2 = ( x + 2y )2
b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\)
c) x3 - 6x2 + 12x - 8 = x3 - 3.x2.2 + 3.x.22 - 23 = ( x - 2 )3
d) x2 - x - y2 - y = ( x2 - y2 ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 )
e) 5x - 5y + ax - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a )
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a) = 5( x2 - 9y2 - 6y - 1 ) = 5[ x2 - ( 9y2 + 6y + 1 ) ] = 5[ x2 - ( 3y + 1 )2 ] = 5( x - 3y - 1 )( x + 3y + 1 )
b) = 125x3 - 25x2 + 15x2 - 3x + 5x - 1 = 25x2( 5x - 1 ) + 3x( 5x - 1 ) + ( 5x - 1 ) = ( 5x - 1 )( 25x2 + 3x + 1 )
c) = 5( x - 7 ) + a( x - 7 ) = ( x - 7 )( a + 5 )
d) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )
e) = ax2 + a - a2x - x = ax( a - x ) + ( a - x ) = ( a - x )( ax + 1 )
f) = ( 10x )2 - ( x2 + 25 )2 = ( 10x - x2 - 25 )( 10x + x2 + 25 ) = -( x - 5 )2( x + 5 )2
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Trả lời:
a, \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)
Vậy x = 0; x = - 1/5 là nghiệm của pt.
b, \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy x = 5 là nghiệm của pt.
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b) \(x^4y^4+64\)
\(=\left(x^4y^4+16x^2y^2+64\right)-16x^2y^2\)
\(=\left(x^2y^2+8\right)^2-\left(4xy\right)^2\)
\(=\left(x^2y^2-4xy+8\right)\left(x^2y^2+4xy+8\right)\)
c) \(x^4y^4+4\)
\(=\left(x^4y^4+4x^2y^2+4\right)-4x^2y^2\)
\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)
a) Áp dụng hằng đẳng thức thứ hai ta có
\(x^2 - 2xy + y^2\) \(<=>\) \((x +y )^2\)
b) Ta có :
\(x^2 - 2xy - 4z^2 + y^2 <=> (x^2 - 2xy + y^2) - (2z)^2 <=> ( x-y)^2 - (2z)^2 <=> ( x-y+2z) (x-y-2z)\)