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\(a,3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\\ b,3x^2+5y-3xy-5x=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\\ c,3y^2-3z^2+3x^2+6xyz=3\left(y^2-z^2+x^2+2xyz\right)\\ d,x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
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a) Ta có: \(3x^2+5y-3xy-5x\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) Ta có: \(3y^2-3z^2+3x^2+6xy\)
\(=3\left(y^2-z^2+x^2+2xy\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
c) Ta có: \(x^2-25-2xy+y^2\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
d) Ta có: \(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
e) Ta có: \(x^2-5x+5y-y^2\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f) Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
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`a)7x^3y^2+14x^2y^3+7xy^4`
`=7xy^2(x^2+2xy+y^2)`
`=7xy^2(x+y)^2`
______________________________________________
`b)x^2-xy+5x-5y`
`=x(x-y)+5(x-y)`
`=(x-y)(x+5)`
______________________________________________
`c)3x^2-6xy-12+3y^2`
`=3(x^2-2xy-4+y^2)`
`=3[(x-y)^2-4]`
`=3(x-y-2)(x-y+2)`
a)7x3y2+14x2y3+7xy4
=7xy2(x2+2xy+y2)
=7xy2(x+y)2
b)x2-xy + 5x - 5y
=x(x-y) + 5(x-y)
=(x-y) (x+5)
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a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
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a) Đề sai nha bạn :) mấy dấu cộng bạn phỉa chuyển thành dấu nhân nhé
\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)
\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(A=2^{512}-1+1\)
\(A=2^{512}\)
b . ( 5x - 3y + 4z )( 5x - 3y - 4z ) = ( 5x - 3y )^2 - ( 4z )^2 = 25x^2 - 30xy + 9y^2 - 16z^2 = 25( y^2 + z^2 ) - 30xy + 9y^2 - 16z^2 = 9z^2 + 34y^2 - 30xy ( 1 )
( 3x - 5y )^2 = 9x^2 - 30xy + 25y^2 = 9( y^2 + z^2 ) - 30xy + 25y^2 = 34y^2 + 9z^2 - 30xy ( 2 )
Tu ( 1 ) va ( 2 ) => dpcm
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\(\frac{5x^2+10xy+5y^2}{3x^3+3y^3}=\frac{5\left(x^2+2xy+y^2\right)}{3\left(x^3+y^3\right)}=\frac{5\left(x+y\right)^2}{3\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{5\left(x+y\right)}{3\left(x^2-xy+y^2\right)}\)
\(\frac{-15x\left(x-y\right)}{3\left(y-x\right)}=\frac{15x\left(y-x\right)}{3\left(y-x\right)}=\frac{15x}{3}\)
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a)\(\frac{12x^5y^2}{8x^3y^5}=\frac{3x^2}{2y^3}\)
b)\(\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
a/ \(\frac{3x^2}{2y^3}\)
b/ \(\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
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a)\(\frac{12x^5y^2}{8x^3y^5}\)
=\(\frac{3x^2}{2y^3}\)
b)\(\frac{x^2+2x+1}{5x^2+5x}\)
=\(\frac{\left(x+1\right)^2}{5x\left(x+1\right)}\)
=\(\frac{x+1}{5x}\)
t tôi nhé bn
\(a,\frac{12x^5y^2}{8x^3y^5}=\frac{4x^3y^2.3x^2}{4x^3y^2.2y^3}=\frac{3x^2}{2y^3}\)
\(b,\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{\left(x+1\right)\left(x+1\right)}{5x\left(x+1\right)}=\frac{x+1}{5x}\)
Xong rồi nhé! Chúc bạn học giỏi.
\(=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}\\ =\dfrac{5}{3}\)
gút wa con ơi