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\(\begin{array}{l}\left( {7{y^5}{z^2} - 14{y^4}{z^3} + 2,1{y^3}{z^4}} \right):\left( { - 7{y^3}{z^2}} \right)\\ = 7{y^5}{z^2}:\left( { - 7{y^3}{z^2}} \right) - 14{y^4}{z^3}:\left( { - 7{y^3}{z^2}} \right) + 2,1{y^3}{z^4}:\left( { - 7{y^3}{z^2}} \right)\\ = - {y^2} + 2yz - 0,3{z^2}\end{array}\)
Bài 1.
x = 14
=> 13 = x - 1 ; 15 = x + 1 ; 16 = x + 2 ; 29 = 2x + 1
Thế vào N(x) ta được :
x5 - ( x + 1 )x4 + ( x + 2 )x3 - ( 2x + 1 )x2 + ( x - 1 )x
= x5 - x5 - x4 + x4 + 2x3 - 2x3 - x2 + x2 - x
= -x = -14
Bài 2.
a) ( 1 - x - 2x3 + 3x2 )( 1 - x + 2x3 - 3x2 )
= [ ( 1 - x ) - ( 2x3 - 3x2 ) ][ ( 1 - x ) + ( 2x3 - 3x2 ) ]
= ( 1 - x )2 - ( 2x3 - 3x2 )2
= 1 - 2x + x2 - [ ( 2x3 )2 - 2.2x3.3x2 + ( 3x2 )2 ]
= x2 - 2x + 1 - ( 4x6 - 12x5 + 9x4 )
= x2 - 2x + 1 - 4x6 + 12x5 - 9x4
= -4x6 + 12x5 - 9x4 + x2 - 2x + 1
b) ( x - y + z )2 + ( z - y )2 + 2( x - y + z )( y - z )
= ( x - y + z )2 + ( z - y )2 - 2( x - y + z )( z - y )
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
\(A=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-z\right)\left(y-x\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
Phân tích tử thức ta có:
\(TS=x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\)
\(=x^2\left(z-y\right)-y^2\left[\left(z-y\right)+\left(y-x\right)\right]+z^2\left(y-x\right)\)
\(=x^2\left(z-y\right)-y^2\left(z-y\right)-y^2\left(y-x\right)+z^2\left(y-x\right)\)
\(=\left(z-y\right)\left(x^2-y^2\right)+\left(y-x\right)\left(z^2-y^2\right)\)
\(=\left(z-y\right)\left(x-y\right)\left(x+y\right)+\left(y-x\right)\left(z-y\right)\left(z+y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(-x-y+z+y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
Vậy \(A=1\)
`8x^4y^5z^3 : 2x^3y^4z`
`= 4xyz^2`.