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a) \(20a^4b^5c^2:\left(-5ab^2c\right)^2=20a^4b^5c^2:\left(25a^2b^4c^2\right)=\dfrac{4}{5}a^2b\)
b) \(\left(-15x^2y^3\right)^7:\left(15xy^3\right)^6-\left(32x^{18}y^5\right):\left(-4x^5y\right)^2=-15x^8y^3-2x^8y^3=-17x^8y^3\)
c) \(-13-13x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)=-13+\dfrac{13}{2}x^4y-\left(x+1\right)^2:\left(x+1\right)=-13+\dfrac{13}{2}x^4y-x-1=-14+\dfrac{13}{2}x^4y-x\)
a: \(\dfrac{20a^4b^5c^2}{\left(-5ab^2c\right)^2}=\dfrac{20a^4b^5c^2}{25a^2b^4c^2}=\dfrac{4}{5}a^2b\)
b: \(\dfrac{\left(-15x^2y^3\right)^7}{\left(15xy^3\right)^6}-\dfrac{\left(32x^{18}y^5\right)}{\left(-4x^5y\right)^2}\)
\(=\dfrac{\left(-15\right)^7\cdot x^{14}\cdot y^{21}}{15^6\cdot x^6\cdot y^{18}}-\dfrac{32x^{18}y^5}{16x^{10}y^2}\)
\(=-15x^8y^3-2x^8y^3\)
\(=-17x^8y^3\)
`-1/3x^5y^2:(-2xy)-(x^2+2x+1):(x+1)`
`=-1/3:(-2).(x^5:x).(y^2:y)-(x+1)^2:(x+1)`
`=-1/6x^4y-(x+1)`
`=-1/6x^4y-x-1`
\(\dfrac{-1}{3}x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)\)
\(=\dfrac{1}{6}x^4y-x-1\)
Lời giải:
a) (x2 + 2xy + y2) : (x + y)
= (x + y)2 : (x + y)
= x + y
b) (125x3 + 1) : (5x + 1)
= [(5x)3 + 1] : (5x + 1)
= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)
= (5x)2 – 5x + 1
= 25x2 – 5x + 1
c) (x2 – 2xy + y2) : (y – x)
= (x – y)2 : [-(x – y)]
= -(x – y)
= y – x
Hoặc (x2 – 2xy + y2) : (y – x)
= (y2 – 2yx + x2) : (y – x)
= (y – x)2 : (y – x)
= y – x
\(a,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x\left(x+5\right)}{x\left(x+6\right)}=\dfrac{x+5}{x+6}\\ b,=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
\(c,=\dfrac{3x^2+2x+1+x^2-2x+1-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)
a) Ta có: ( x2 -1 )( x2 + 2x )
= x2( x2 + 2x ) - ( x2 + 2x )
= x4 + 2x3 - x2 - 2x
b) Ta có ( x + 3 )( x2 + 3x -5 )
= x( x2 + 3x -5 ) + 3( x2 + 3x -5 )
= x3 + 3x2 - 5x + 3x2 + 9x - 15
= x3 + 6x2 + 4x - 15
c) Ta có ( x -2y )( x2y2 - xy + 2y )
= x( x2y2 - xy + 2y ) - 2y( x2y2 - xy + 2y )
= x3y2 - x2y + 2xy - 2x2y3 + 2xy2 - 4y2
d) Ta có ( 1/2xy -1 )( x3 -2x -6 )
= 1/2xy( x3 -2x -6 ) - ( x3 -2x -6 )
= 1/2x4y - x2y - 3xy - x3 + 2x + 6
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
a. 5x + 3(x2 - x - 1)
= 5x + 3x2 - 3x - 3
= 3x2 + 5x - 3x - 3
= 3x2 + 2x - 3
b. (5 - x)(5 + x) - (2x - 1)2
25 - x2 - (4x2 - 4x + 1)
= 25 - x2 - 4x2 + 4x - 1
= 25 - 1 - x2 - 4x2 + 4x
= 24 - 5x2 + 4x
\(a,=4x^2+12xy+9y^2\\ b,=25x^2-10xy+y^2\\ d,=4x^2+4xy^2+y^4\\ e,=9x^4-12x^2y+4y^2\\ g,=x^3+64\)
a: \(\dfrac{20a^4b^5c^2}{\left(-5ab^2c\right)^2}\)
\(=\dfrac{20a^4b^5c^2}{25a^2b^4c^2}\)
\(=\dfrac{4}{5}a^2b\)
b: \(\dfrac{\left(-15x^2y^3\right)^7}{\left(15xy^3\right)^6}-\dfrac{32x^{18}y^5}{\left(-4x^5y\right)^2}\)
\(=\dfrac{-15^7\cdot x^{14}\cdot y^{21}}{15^6\cdot x^6\cdot y^{18}}-\dfrac{32x^{18}y^5}{16x^{10}y^2}\)
\(=-15x^8y^3-2x^8y^3\)
c: \(\dfrac{-\dfrac{1}{3}x^5y^2}{-2xy}-\dfrac{x^2+2x+1}{x+1}\)
\(=\dfrac{2}{3}x^3y-x-1\)