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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(x^2-5y+y^2-2xy+5x=\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\) a/ x2 – 5y + y2 -2xy + 5x = ( x2 - 2xy + y2 ) - 5( y - x ) = ( x - y )2 - 5( y - x ) = ( y - x )2 - 5( y - x ) = ( y - x )( y - x - 5 ) b/ 4x2 – 81(y – 2)2 = 4x2 - 92(y – 2)2= 4x2 – ( 9y – 18)2 = ( 2x -9y -18 )( 2x + 9y + 18 ) c/ x2z – y2z + 2yz – z = ( x2z + yz ) - ( y2z - yz ) - z = z( x2 + y ) - z( y2 - y ) -z = z( x2 + y - y2 +y - 1 ) = z( x2 + 2y - y2 - 1 ) \(=z[x^2-\left(y^2-2y+1\right)]=z[x^2-\left(y-1\right)^2=z\left(x-y+1\right)\left(x+y-1\right)\) d/ x3 – 8y3 + x2 + 2xy + 4y2 = ( x3 – 8y3 ) + x2 + 2xy + 4y2 = ( x -2y )( x2 + 2xy + 4y2 ) + ( x2 + 2xy + 4y2 0 = ( x2 + 2xy + 4y2)( x -2y +1) e/ 7x2 – 11x + 4 = 7x2 -7x -4x +4 = 7x( x-1 ) - 4( x - 1 ) = ( x - 1 )( 7x - 4 ) g/ 13x2 + 2xy – 15y2 = 13x2 - 13xy + 15xy - 15y2 = 13x( x - y ) + 15y( x - y ) = ( x - y )( 13x + 15y ) h/ x3 + 3x2 + 3x + 2 = x3 +2x2 + x2 +2x + x + 2 = x2( x + 2 ) + x( x + 2 ) + ( x + 2 ) = ( x + 2 )( x2 + x + 1 ) i/ x3 – 3x2 + 3x – 2 + xy – 2y = x3 - 2x2 - x2 + 2x + x - 2 +xy - 2y = x2( x - 2 ) - x( x - 2 ) + ( x - 2 ) + y( x - 2 ) = ( x - 2 )( x2 - x +1 + y ) Trả lời: a) x2 + 4y2 + 4xy = x2 + 2.x.2y + (2y)2 = ( x + 2y )2 b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\) c) x3 - 6x2 + 12x - 8 = x3 - 3.x2.2 + 3.x.22 - 23 = ( x - 2 )3 d) x2 - x - y2 - y = ( x2 - y2 ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 ) e) 5x - 5y + ax - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a ) g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\) f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\) e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\) d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\) Answer: \(5x^2-10xy+5y^2-20z^2\) \(=5.\left(x^2-2xy+y^2-4z^2\right)\) \(=5.[\left(x+y\right)^2-\left(2z\right)^2]\) \(=5.\left(x+y-2z\right).\left(x+y+2z\right)\) \(16x-5x^2-3\) \(=\left(-5x^2+15x\right)+\left(x-3\right)\) \(=-5x.\left(x-3\right)+\left(x-3\right)\) \(=\left(1-5x\right).\left(x-3\right)\) \(x^2-5x+5y-y^2\) \(=(x-y).(x+y)-5.(x-y)\) \(=(x-y).(x+y-5)\) \(3x^2-6xy+3y^2-12z^2\) \(=3.(x^2-2xy+y^2-4z^2)\) \(=3[\left(x-y\right)^2-\left(2z\right)^2]\) \(=3.(x-y-2z).(x-y+2z)\) \(x^2+4x+3\) \(=(x^2+x)+(3x+3)\) \(=x.(x+1)+3.(x+1)\) \(=(x+1).(x+3)\) \((x^2+1)^2-4x^2\) \(=(x^2-2x+1).(x^2+2x+1)\) \(=(x-1)^2.(x+1)^2\) \(x^2-4x-5\) \(=(x^2+x)-(5x+5)\) \(=x.(x+1)-5.(x+1)\) \(=(x-5).(x+1)\) Bài 14:Tìm x a,\(x-3=\left(3-x\right)^2\) \(\Rightarrow\left(x-3\right)-\left(3-x\right)^2=0\) \(\Rightarrow\left(x-3\right)+\left(x-3\right)^2=0\) \(\Rightarrow\left(x-3\right)\left(1+x-3\right)=0\) \(\Rightarrow\left(x-3\right)\left(x-2\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\) b,\(\left(2x-5\right)-\left(5+2x\right)^2=0\) \(\Rightarrow\left(2x-5\right)+\left(2x-5\right)^2=0\) \(\Rightarrow\left(2x-5\right)\left(1+2x-5\right)=0\) \(\Rightarrow\left(2x-5\right)\left(2x-4\right)=0\) \(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=5\\2x=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
K
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CM
Cao Minh Tâm
28 tháng 8 2019
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