YS
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
T
24 tháng 4 2019
b) Dùng phương pháp đặt ẩn phụ:
Đặt y - x = a; z - y = b suy ra \(a+b=y-x+z-y=z-x\)
\(x^2y^2a+y^2z^2b-z^2x^2\left(a+b\right)=\left(x^2y^2a-z^2x^2a\right)+\left(y^2z^2b-z^2x^2b\right)\)
\(=x^2a\left(y^2-z^2\right)+z^2b\left(y^2-x^2\right)=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)+z^2\left(z-y\right)\left(y-x\right)\left(x+y\right)\)
\(=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)-z^2\left(y-z\right)\left(y-x\right)\left(x+y\right)\)
\(=\left(y-x\right)\left(y-z\right)\left[x^2\left(y+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(y-x\right)\left(y-z\right)\left(x^2y+x^2z-z^2x-z^2y\right)\)
\(=\left(y-x\right)\left(y-z\right)\left[y\left(x^2-z^2\right)+xz\left(x-z\right)\right]\)
\(=\left(y-x\right)\left(y-z\right)\left[y\left(x-z\right)\left(x+z\right)+xz\left(x-z\right)\right]\)
\(=\left(y-x\right)\left(y-z\right)\left(x-z\right)\left(xy+yz+zx\right)\)
8 tháng 10 2018
\(a)\)\(\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\)\(\left(x^2+y^2-5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\)\(\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\)\(\left(x^2-2xy+y^2-5+4\right)\left(x^2+2xy+y^2-5-4\right)\)
\(=\)\(\left[\left(x-y\right)^2-1\right].\left[\left(x+y\right)^2-9\right]\)
\(=\)\(\left(x-y-1\right)\left(x-y+1\right)\left(x+y-9\right)\left(x+y+9\right)\)
Chúc bạn học tốt ~
31 tháng 7 2018
b) (1 + 2x)(1- 2x) - x(x+2)(x-2)
= (1- 4x2) - x(x2 - 4)
= 1 - 4x2- x3- 4x
= (1 - x3) + (4x - 4x2)
= (1- x) (1 + x + x2) + 4x(1 -x)
= (1-x)(1+5x + x2)
YS
1
FM
16 tháng 10 2018
Ta có:
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x-y+x\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-x\right)-y^2z^2\left(y-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)+z^2\left(z-x\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y^2x+y^2z-z^2y-z^2x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)
16 tháng 7 2019
\(\left(2a+b\right)^2-\left(2a+a\right)^2\)
\(=\left(2a+b-2a-a\right)\left(2a+b+2a+a\right)\)
\(=\left(b-a\right)\left(5a+b\right)\)
16 tháng 7 2019
\(\left(2a+b\right)^2-\left(2a+a\right)^2\)
\(=\left(2a+b\right)^2-\left(3a\right)^2\)
\(=\left(2a+b-3a\right)\left(2a+b+3a\right)\)
\(=\left(b-a\right)\left(5a+b\right)\)
VM
1
MT
30 tháng 9 2015
a) x3+y3+z3-3xyz
=(x+y)3+z3-3x2y-3xy2-3xyz
=(x+y+z).[(x+y)2+(x+y).z+z2]-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2)-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2-3xy)
=(x+y+z)(x2+y2+zx+zy+z2-zy)
b)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(-a2c+c2a)+(b2c-b2a)
=b.(a2-c2)-ac.(a-c)-b2.(a-c)
=b.(a+c)(a-c)-ac.(a-c)-b2.(a-c)
=(a-c)[b.(a+c)-ac-b2]
=(a-c)(ab+bc-ac-b2)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
20 tháng 9 2015
a/ (x-a)^4 - (x+a)^4
=((x-a)^2)^2 - ((x+a)^2)^2
=(x^2 - 2xa + a^2)^2 - (x^2 +2xa+a^2)^2
=(x^2-2xa+a^2-x^2-2xa-a^2)(x^2-2xa+a^2+x^2+2xa+a^2)
=-4xa(2x^2+2a^2)
b/ x^4 –y^2(2x-y)^2
=(x^2)^2-(y(2x-y)^2
=(x^2)^2-(2xy-y^2)^2
=(x^2-2xy+y^2)(x^2+2xy+y^2)
=(x-y)^2 (x+y)^2
c/(xy+4)^2- 4(x+y)^2
=(xy+4)^2- (2x+2y)^2
=(xy+y-2x-2y)(xy+y+2x+2y)
=(xy-y+2x)(xy+3y+2x)
NN
15 tháng 10 2020
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
G
15 tháng 10 2020
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(16^4+y^4=\left[\left(y^2\right)^2+2.y^2.16^2+\left(16^2\right)^2\right]-2.y^2.16^2=\left(y^2+16^2\right)^2-2.y^2.16^2\)
b tự tính tiếp nhé
ý b tương tự. ( gợi ý: thêm bớt hạng tử 16y^4 )
\(y^8+64\)
\(=\left(y^4\right)^2+2\cdot y^4\cdot8+8^2-2\cdot y^4\cdot8\)
\(=\left(y^4+8\right)^2-16y^4\)
\(=\left(y^4+8\right)^2-\left(4y^2\right)^2\)
\(=\left(y^4+8-4y^2\right)\left(y^4+8+4y^2\right)\)
a kudo shinichi làm rồi đó