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26 tháng 7 2020
a) ( x2 - 5 )( x + 3 ) = x3 + 3x2 - 5x - 15
b) ( x + 4 )( x - x2 ) = x2 - x3 + 4x - 4x2 = -x3 - 3x2 + 4x
c) ( x2 - 6 )( x + 2 ) + ( x + 3 )( x - x2 ) = x3 + 2x2 - 6x - 12 + x2 - x3 + 3x - 3x2 = -3x - 12 = -3( x + 4 )
d) x( x - y ) - y( x - y ) = ( x - y )( x - y ) = ( x - y )2
e) x2( x + y ) - x( x2 - y ) = x3 + x2y - x3 + xy = x2y + xy = xy( x + 1 )
f) 3x( 12x - 4 ) - 9x( 4x - 3 ) = 36x2 - 12x - 36x2 + 27x = 15x
26 tháng 7 2020
Bài làm
a) ( x2 - 5 )( x + 3 )
= x3 + 3x2 - 5x - 15
b) ( x + 4 )( x - x2 )
= ( x + 4 ) . x( 1 - x )
= x( x + 4 )( 1 - x )
= x( x - x2 + 4 - 4x )
= x( 4 - x2 - 3x )
= 4x - x3 - 3x2
c) ( x2 - 6 )( x + 2 ) + ( x + 3 )( x - x2 )
= ( x - 3 )( x + 3 )( x + 2 ) + ( x + 3 )( x - x2 )
= ( x + 3 )[ ( x - 3 )( x + 2 ) + ( x - x2 )]
= ( x + 3 ) [ x2 + 2x - 3x - 6 + x2 - x2 ]
= ( x + 3 ) ( x2 - x - 6 )
= x3 - x2 - 6x + 3x2 - 3x - 18
= x3 + 2x2 - 9x - 18
d) x( x - y ) - y( x - y )
= ( x - y )( x - y )
= ( x - y )2
= x2 - 2xy + y
e) x2( x + y ) - x( x2 - y )
= x3 + x2y - x3 + xy
= x2y + xy
f) 3x( 12x - 4 ) - 9x( 4x - 3 )
= 3x . 3( 4x - 1 ) - 9x( 4x - 3 )
= 9x( 4x - 1 ) - 9x( 4x - 3 )
= 9x( 4x - 1 - 4x + 3 )
= 9x . 2
= 18x
19 tháng 8 2018
https://olm.vn/thanhvien/quynhgiang2k4 à mình quên ghi đề bài là:
rút gọn biểu thức nha
TN
6 tháng 6 2017
\(a,\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2\left(x^2+y^2\right)\)\(b,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2=2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3x^2\)\(c,\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2=\left(x-2y\right)^2\)
17 tháng 6 2017
a) \(\left(x+y\right)^2+\left(x-y\right)^2\)
=\(\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)\)
=\(x^2+2xy+y^2+x^2-2xy+y^2\)
\(2x^2+2y^2=2\left(x^2+y^2\right)\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left[\left(x-y\right)+\left(x+y\right)\right]^2\)
= \(\left(x-y+x+y\right)^2\)
\(=2x^2\)
c) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2-2\left(x-y+z\right)\left(z-y\right)+\left(z-y\right)^2\)
\(=\left[\left(x-y+z\right)-\left(z-y\right)\right]^2\)
= \(\left(x-y+z-z+y\right)^2=x^2\)
DH
5 tháng 7 2017
a, \(\left(x+3y\right)\left(x^2-2xy+y\right)=x^3-2x^2y+xy+3x^2y-6xy^2+3y^2\)
\(=x^3+x^2y+xy-6xy^2+3y^2\)
\(b,\left(x^2-1\right)\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
c, \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x=17x^2-6x^3+5x-6\)
d, \(\left(x+2\right)\left(x-1\right)=x^2-x+2x-2=x^2+x-2\)
e, \(x\left(x-y\right)-y\left(y-x\right)=x^2-xy-\left(y^2-xy\right)\)
\(=x^2-xy-y^2+xy=x^2-y^2\)
Chúc bạn học tốt!!!
TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
2 tháng 9 2017
a ) ( x + y )2 +( x - y )2 = x2 + 2xy +y2 + x2 - 2xy + y2
= 2x2 + 2y2
b ) 2 . ( x - y ) . ( x + y ) + ( x + y )2 + ( x - y )2
= 2 . ( x2 - y2 ) + x2 + 2xy + y2 + x2 - 2xy + y2
= 2x2 - 2y2 + x2 +2xy + y2 + x2 - 2xy + y2
= 4x2
c ) ( x - y + z )2 - ( z - y )2 + 2.( x - y + z ) ( y - z )
= x2 + y2 + z2 - 2xy + 2 xz - 2yz - z2 + 2zy - y2 + 2xy - y2 + 2yz -2xz + 2y2 - 2z2
= x2
\(x\left(x^2-y\right)+x^2\left(x+y\right)\)
\(=xx^2-xy+x^2x+x^2y\)
\(=x^3-xy+x^3+x^2y\)
\(=2x^3-xy+x^2y\)
Bài làm:
Ta có: \(x\left(x^2-y\right)+x^2\left(x+y\right)\)
\(=x^3-xy+x^3+x^2y\)
\(=x^2y-xy\)
\(=xy\left(x-1\right)\)(nếu PTĐTTNT)