Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)x^2.16-4xy+4y^2
<=>16.x^2-2x2y+(2y)^2
<=>16(x-2y)^2
b)x^5-x^4+x^3-x^2
<=>(x^5-x^4)+(x^3-x^2)
<=>x^4(x-1)+x^2(x-1)
<=>(x-1)(x^4+x^2)
c)x^5+x^3-x^2-1
<=>(x^5+x^3)-(x^2+1)
<=>x^3(x^2+1)-(x^2+1)
<=>(x^2+1)(x^3-1)
d)x^4-3x^3-x+3
<=>(x^4-3x^3)-(x-3)
<=>x^3(x-3)-(x_3)
<=>(x-3)(x^3-1)
\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)
a)\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
b)\(x^4-x^3-x^2+1=\left(x^4-x^3\right)-\left(x^2-1\right)=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
c)\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
a. 2x-1-x2= -(x2-2x+1)=-(x-1)2
b. 8x3+y6=(2x)3+(y2)3
=(2x+y2)(4x2-2xy2+y4)
c. x2-16+4xy+4y2=(x2+4xy+4y2)-16
=(x+2y)2-16=(x+2y+4)(x+2y-4)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) 16(4x+5)2 - 25(2x+2)2
\(=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2\)
\(=\left[4\left(4x+5\right)+5\left(2x+2\right)\right]\left[4\left(4x+5\right)-5\left(2x+2\right)\right]\)
\(=\left(16x+20+10x+10\right)\left(16x+20-10x-10\right)\)
\(=\left(26x+30\right)\left(6x+10\right)\)
\(b,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-2y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-3y+5\right)\)
\(c,\left(x+1\right)^4-\left(x-1\right)^4\)
\(=\left(x+1\right)^{2^2}-\left(x-1\right)^{2^2}\)
\(=\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\)
\(=\left(x^2+2x+1+x^2-2x+1\right)\left[\left(x+1+x-1\right)\left(x+1-x+1\right)\right]\)
\(=\left(2x^2+2\right)2x.2\)
\(=4x.2\left(x^2+1\right)\)
\(=8x\left(x^2+1\right)\)
\(x^3-2x^2-4xy^2+x\)
\(=x\left(x^2-2x-4y^2+1\right)\)
\(=x\left(\left(x-1\right)^2-\left(2y\right)^2\right)\)
\(=x\left(x-1-2y\right)\left(x-1+2y\right)\)
1) \(25x^4-10x^2y+y^2\)
\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)
\(\Leftrightarrow\left(5x^2+y\right)^2\)
2) \(x^4+2x^3-4x-4\)
\(\Leftrightarrow\left(x^4-4\right)+\left(2x^3-4x\right)\Leftrightarrow\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2+2x\right)\)
3) \(x^4+x^2+1\)
\(\Leftrightarrow x^4+x^2-x+x+1\)
\(\Leftrightarrow\left(x^4-x\right)+\left(x^2+x+1\right)\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)
\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)
5) \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)
\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)
\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\)
\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)
\(\left(4-x\right)^2+\left(x-4\right)\left(x-5\right)-4\left(x-5\right)^2+1\)
= \(16-4x+x^2+x^2-5x-4x+20-4\left(x^2-5x+25\right)+1\)
= \(37-13x+2x^2-4x^2+20x+100\)
= \(137+7x-2x^2\)
\(=\left(x-4\right)^2+\left(x-4\right)\left(x-5\right)-\left(2\left(x-5\right)\right)^2+1\)
\(=\left(x-4\right)\left(2x-9\right)-\left(\left(2x-10\right)^2-1\right)\)
\(=\left(x-4\right)\left(2x-9\right)-\left(2x-11\right)\left(2x-9\right)\)
\(=\left(2x-9\right)\left(x-4-2x+11\right)=\left(2x-9\right)\left(7-x\right)\)
a) x2 - 16 - 4xy + 4y2
= ( x2 - 4xy + 4y2 ) - 16
= ( x - 2y )2 - 42
= ( x - 2y - 4 )( x - 2y + 4 )
b) x5 - x4 + x3 - x2
= x2( x3 - x2 + x - 1 )
= x2[ x2( x - 1 ) + ( x - 1 ) ]
= x2( x - 1 )( x2 + 1 )
c) x( x + 4 )( x + 6 )( x + 10 ) + 128 < mình nghĩ là nên sửa đề như này :]>
= [ x( x + 10 ) ][ ( x + 4 )( x + 6 ) ] + 128
= ( x2 + 10x )( x2 + 10x + 24 ) + 128
Đặt t = x2 + 10x
bthuc <=> t( t + 24 ) + 128
= t2 + 24t + 128
= t2 + 16t + 8t + 128
= t( t + 16 ) + 8( t + 16 )
= ( t + 16 )( t + 8 )
= ( x2 + 10x + 16 )( x2 + 10x + 8 )
= ( x2 + 2x + 8x + 16 )( x2 + 10x + 8 )
= [ x( x + 2 ) + 8( x + 2 ) ]( x2 + 10x + 8 )
= ( x + 2 )( x + 8 )( x2 + 10x + 8 )
cảm ơn bạn câu c mình chép nhầm nó là 128 đó