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a, \(=-91x-y+5z\)
b, \(=4x^2+x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2+x^2y\)
\(=4x^2+2x^2y-5y^2-\dfrac{5}{3}x^3+6xy^2\)
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
Bài 1:
a) Ta có: \(2x=5y.\)
=> \(\frac{x}{y}=\frac{5}{2}\)
=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)
Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)
Có: \(x.y=90\)
=> \(5k.2k=90\)
=> \(10k^2=90\)
=> \(k^2=90:10\)
=> \(k^2=9\)
=> \(k=\pm3.\)
TH1: \(k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)
TH2: \(k=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)
e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)
=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)
Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)
Có: \(x.y=20\)
=> \(4k.5k=20\)
=> \(20k^2=20\)
=> \(k^2=20:20\)
=> \(k^2=1\)
=> \(k=\pm1.\)
TH1: \(k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)
TH2: \(k=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)
Chúc bạn học tốt!
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: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)
\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)
\(=-15x^3y^2+25x^2y^2-5xy^3\)
1 + 2xy - x2 - y2
= 1 - ( x2 - 2xy + y2 )
= 12 - ( x - y )2
= [ 1 - ( x - y ) ][ 1 + ( x - y ) ]
= ( y - x + 1 )( x - y + 1 )
a2 + b2 - c2 - d2 - 2ab + 2cd
= ( a2 - 2ab + b2 ) - ( c2 - 2cd + d2 )
= ( a - b )2 - ( c - d )2
= [ ( a - b ) - ( c - d ) ][ ( a - b ) + ( c - d ) ]
= ( a - b - c + d )( a - b + c - d )
a3b3 - 1
= ( ab )3 - 13
= ( ab - 1 )[ ( ab )2 + ab.1 + 12 ]
= ( ab - 1 )( a2b2 + ab + 1 )
x2( y - z ) + y2( z - x ) + z2( x - y )
= z2( x - y ) + x2y - x2z + y2z + y2x
= z2( x - y ) + ( x2y - y2x ) - ( x2z - y2z )
= z2( x - y ) + xy( x - y ) - z( x2 - y2 )
= z2( x - y ) + xy( x - y ) - z( x + y )( x - y )
= ( x - y )[ z2 + xy - z( x + y ) ]
= ( x - y )( z2 + xy - zx - zy )
= ( x - y )[ ( z2 - zx ) - ( zy - xy ) ]
= ( x - y )[ z( z - x ) - y( z - x ) ]
= ( x - y )( z - x )( z - y )
B1 : a, M = x3-3xy(x-y)-y3-x2+2xy-y2
= ( x3-y3)-3xy(x-y) -(x2-2xy+y2)
= (x-y)(x2+xy+y2)-3xy(x-y)-(x-y)2
= (x-y) [(x2+xy+y2-3xy-(x-y)]
= (x-y)[(x2-2xy+y2)-(x-y)
= (x-y)[(x-y)2-(x-y)]
= (x-y)(x-y)(x-y-1)
= (x-y)2(x-y-1)
= 72(7-1) = 49 . 6= 294
N = x2(x+1)-y2(y-1)+xy-3xy(x-y+1)-95
= x3+x2-(y3-y2)+xy-(3x2y-3xy2+3xy)-95
= x3+x2-y3+y2+xy-3x2y+3xy2-3xy-95
= (x3-y3)+(x2-2xy+y2)-(3x2y+y2)-(3x2y-3xy2)-95
=(x-y)(x2+xy+y2)+(x-y)2-3xy(x-y)-95
= (x-y)(x2+xy+y2+x-y-3xy)-95
= (x-y)[(x2-2xy+y2)+(x-y)]-95
= (x-y)[(x-y)2+(x-y)]-95
=(x-y)(x-y)(x-y+1)-95
= (x-y)2(x-y+1)-95
= 72(7+1)-95=297
1: \(=-\left(x^2+2x+2\right)=-\left(x^2+2x+1+1\right)=-\left(x+1\right)^2-1< =-1\)
Dấu '=' xảy ra khi x=-1
2: \(=-\left(4x^2-12x-10\right)\)
\(=-\left(4x^2-12x+9-19\right)\)
\(=-\left(2x-3\right)^2+19< =19\)
Dấu '=' xảy ra khi x=3/2
3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)
Dấu '=' xảy ra khi x=-2
