Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
Bài 8:
b. 1+8x6y3 = 13+23(x2)3y3 = 13+(2x2y)3
= (1+2x2y)(1-2x2y+4x4y2)
e. 27x3+\(\dfrac{y^3}{8}\)\(=\left(3x\right)^3+\left(\dfrac{y}{2}\right)^3\)
= (3x+\(\dfrac{y}{2}\))(9x2-\(\dfrac{3xy}{2}\)+\(\dfrac{y^2}{4}\))
Bài 9:
c. 1- 9x +27x2 -27x3 = 13-3.12.3x+3.(3x)2-(3x)3
= (1-3x)3
d. x3+\(\dfrac{3}{2}x^2\)+\(\dfrac{3}{4}x+\dfrac{1}{8}\) = x3+\(3x^2.\dfrac{1}{2}\)+\(3x.\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3\)
= (x+\(\dfrac{1}{2}\))3
f. x2 - 2xy +y2 -4m2 +4m.n - n2 = (x2 - 2xy +y2)-((2m)2 -2.2m.n + n2)
= (x-y)2-(2m-n)2 = (x-y-2m+n)(x-y+2m-n)
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
a, (2xy+1)2-(2x+y)2=(2xy+1-2x-y)(2xy+1+x+y)=[(2xy-2x)-(y-1)][(2xy+2x)-(y+1)]=[2x(y-1)-(y-1)][2x(y+1) +(y+1)]
=(y-a)(2x-1)(2x+1)(y+1)
b, x2(x-2)2-(x-2)2-x2+1=[x2(x-2)2-(x-2)2] - (x2-1)=(x-2)2(x2-1) - (x2-1)=(x2-1)[(x-2)2-1]=(x-1)(x+1)(x+1)(x-3)
c,x4+2x3-2x-1=(x4-1)+(2x3-2x)=(x2-1)(x2 +1)+2x(x2-1)=(x2-1)(x2+2x+1)=(x-1)(x+1)(x+1)2
d,1+6x-6x2-x3=
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) \(16x^2-\left(x^2+4\right)^2=\left(4x\right)^2-\left(x^2+4\right)^2\)
\(=\left(4x+x^2+4\right)\left(4x-x^2-4\right)\)
\(=\left(x+2\right)^2\left\{-\left(x^2-4x+4\right)\right\}=\left(x+2\right)^2\left\{-\left(x-2\right)^2\right\}\)
Ở đây mình không đổi \(-\left(x-2\right)^2=\left(2-x\right)^2\)được vì vốn dĩ \(\left(x-2\right)^2=\left(2-x\right)^2\)
b) \(\left(x^2+9\right)^2-36=\left(x^2+9\right)^2-6^2\)
\(=\left(x^2+9+6\right)\left(x^2+9-6\right)=\left(x^2+15\right)\left(x^2-3\right)\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
f) \(x^4y^4-z^4=\left\{\left(x^2y^2\right)^2-\left(z^2\right)^2\right\}\)
\(=\left\{\left(xy\right)^2-z^2\right\}\left\{x^2y^2+z^2\right\}\)
\(=\left(xy-z\right)\left(xy+z\right)\left(x^2y^2+x^2\right)\)