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\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
a/ \(x^3-5x^2+8x-4\)
= \(\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
b/ \(x^3-x^2+x-1\)
= \(\left(x^3-x^2\right)+\left(x-1\right)\)
= \(x^2\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+1\right)\)
a) Ta có : x2 - 4x + 3
= x2 - x - 3x + 3
= x(x - 1) - (3x - 3)
= x(x - 1) - 3(x - 1)
= (x - 1) (x - 3)
a) \(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
b) \(x^2+5x+4\)
\(=x^2+x+4x+4\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+4\right)\)
c) \(x^2-x-6\)
\(=x^2-3x+2x-6\)
\(=x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x+2\right)\left(x-3\right)\)
d) \(x^4+1997x^2+1996x+1997\)
\(=x^4+x^2+1996x^2+1996x+1996+1\)
\(=\left(x^4+x^2+1\right)+\left(1996x^2+1996x+1996\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
e) \(x^2-2001\cdot2002\)( hình như sai sai)
a) bt \(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x+1\right)\left(x-2\right)\)
kl: ...
b) \(=\left(x+2\right)\left(x^2-8x-15\right)=\left(x+2\right)\left(x-5\right)\left(x-3\right)\)
kl:....
a, \(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)\)
\(=\left(x-8\right)\left(x^2-2x+x-2\right)\)
\(=\left(x-8\right)\left[x\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
b, \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left(x^2-3x+2x-6\right)\)
\(=\left(x-5\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)
Chúc bạn học tốt!!!
a) x2 + 3x - 18 = 0
⇔ x2 - 3x + 6x - 18 = 0
⇔ x( x - 3 ) + 6( x - 3 ) = 0
⇔ ( x - 3 )( x + 6 ) = 0
⇔ x - 3 = 0 hoặc x + 6 = 0
⇔ x = 3 hoặc x = -6
b) x3 - x2 - 4 = 0
⇔ x3 - 2x2 + x2 - 4 = 0
⇔ x2( x - 2 ) + ( x - 2 )( x + 2 ) = 0
⇔ ( x - 2 )( x2 + x + 2 ) = 0
⇔ x - 2 = 0 hoặc x2 + x + 2 = 0
⇔ x = 2 < do x2 + x + 2 = ( x2 + x + 1/4 ) + 7/4 = ( x + 1/2 )2 + 7/4 ≥ 7/4 > 0 ∀ x
b) x3 - 6x2 - x + 30 = 0
⇔ x3 - 5x2 - x2 + 5x - 6x + 30 = 0
⇔ x2( x - 5 ) - x( x - 5 ) - 6( x - 5 ) = 0
⇔ ( x - 5 )( x2 - x - 6 ) = 0
⇔ ( x - 5 )( x2 - 3x + 2x - 6 ) = 0
⇔ ( x - 5 )[ x( x - 3 ) + 2( x - 3 ) ] = 0
⇔ ( x - 5 )( x - 3 )( x + 2 ) = 0
⇔ x - 5 = 0 hoặc x - 3 = 0 hoặc x + 2 = 0
⇔ x = 5 hoặc x = 3 hoặc x = -2
Thời gian có hạn copy cái này hộ mình vào google xem nha :
https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi
Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....
Có 500 giải nhanh nha đã có 401 người nhận rồi
OKz
a) \(x^3+x^2+5x^2+5x+6x+6=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b) \(x^3-3x^2+9x^2-27x+14x-42\)
\(=x^2\left(x+3\right)+9x\left(x+3\right)+14\left(x+3\right)\)
\(=\left(x^2+9x+14\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(x+2\right)\left(x+7\right)\)
c) \(\left(x^2+x+4\right)^2+3x\left(x^2+x+4\right)+5x\left(x^2+x+4\right)+15x^2\)
\(=\left(x^2+x+4\right)\left(x^2+x+4+3x\right)+5x\left(x^2+x+4+3x\right)\)
\(=\left(x^2+6x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x^2+6x+4\right)\left(x+2\right)^2\)
d) \(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+16.24+16\)
\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+400\)
\(=\left(x^2+10x+20\right)^2\)
a)
\(\left(1-x^3\right)+\left(6x-6x^2\right)=\left(1-x\right)\left[\left(x^2+x+1\right)+6x\right]=\left(1-x\right)\left(x^2+7x+1\right)\)b) check lại đề
c)
(x+4)^2 (x^2 -1) -(x^2 -1) =(x^2 -1)[(x+4)^2 -1]
\(\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)