Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=3x^2\left(x^2-1\right)+\left(x^8-3x^4+3x^2-1\right)-\left(x^8-1\right)\)
\(=3x^4-3x^2+x^8-3x^4+3x^2+1-x^8+1\)
\(=2\)
=2 nha ban
(con cach lam ban nhan dang thuc len rui rut gon lai)
a) = (x+1-x+1)(x2+2x+1+x2-1+x2-2x+1)- 6(x2-1)
= 2( 3x2+1)- 6(x2-1)
= 2( 3x2+1-3x2+3)
=2. 4
=8
\(a,\left(-4xy-5\right)\left(5-4xy\right)=\left(4xy+5\right)\left(4xy-5\right).\)
\(=\left(4xy\right)^2-5^2=16x^2y^2-25\)
\(b,\left(a^2b+ab^2\right)\left(ab^2-a^2b\right)=\left(ab^2+a^2b\right)\left(ab^2-a^2b\right)\)
\(=\left(ab^2\right)^2-\left(a^2b\right)^2=a^2b^4-a^4b^2\)
\(c,\left(3x-4\right)^2+2\left(3x-4\right)\left(4-x\right)+\left(4-x\right)^2\)
\(=\left[\left(3x-4\right)+\left(4-x\right)\right]^2\)
\(=\left(3x-4+4-x\right)^2=\left(2x\right)^2=4x^2\)
\(d,\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)-\left(a^4+b^4\right)\)
\(=\left[\left(a^2+b^2\right)+ab\right]\left[\left(a^2+b^2\right)-ab\right]-\left(a^4+b^4\right)\)
\(=\left(a^2+b^2\right)^2-\left(ab\right)^2-a^4-b^4\)
\(=a^4+2a^2b^2+b^4-a^2b^2-a^4-b^4=a^2b^2\)
\(\left(2x+1\right)\left(x+3\right)+\left(x+1\right)^2\left(x+2\right)+\left(x+5\right)\left(x+1\right)\)
\(=2x^2+6x+x+3+x^3+2x^2+x+2x^2+4x+2+x^2+x+5x+5\)
\(=x^3+7x^2+18x+10\)
đúng ko nhỉ?
tham khảo : KHAI TRIỂN RÚT GỌN ĐA THỨC BẰNG CASIO (1LINK DUY NHẤT) - YouTube
a) \(\left(1+x\right)^2+\left(1-x\right)^2\)
\(=1+2x+x^2+1-2x+x^2\)
\(=2x^2+2\)
b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)
\(=x^2+4x+4+1-x^2\)
\(=4x+5\)
c) \(\left(x-3\right)^2+3\left(x+1\right)^2\)
\(=x^2-6x+9+3x^2+6x+3\)
\(=4x^2+12\)
d)\(\left(2+3x\right)\left(3x-2\right)-\left(3x+1\right)^2\)
\(=9x^2-4-9x^2-6x-1\)
\(=-6x-5\)
e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=x^2-2x+5x-10-x^2-4x-4\)
\(=-x-14\)
f) \(\left(x+3\right)\left(2x-5\right)-2\left(1+x\right)^2\)
\(=2x^2-5x+6x-15-2-4x-2x^2\)
\(=-3x-17\)
g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)
\(=16x^2-1-4+16x-16x^2\)
\(=16x-5\)
#Học tốt!
a,\(\left(2x-1\right)\left(4x^2+2x+1\right)=\left(2x-1\right)\left[\left(2x\right)^2+2x.1+1^2\right]\)
\(=\left(2x\right)^3-1=8x^3-1\)
b,\(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)
\(=x^2+2.x.2y+\left(2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)
`a)(2x-1)(4x^2+2x+1)`
`=(2x-1)[(2x)^2+2x.1+1^2]`
`=(2x)^3-1^3`
`=8x^3-1`
Áp dụng HĐT:`A^3-B^3=(A-B)(A^2+AB+B^2)`
`b)(x+2y+z)(x+2y-z)`
`=[(x+2y)+z][(x+2y)-z]`
`=(x+2y)^2-z^2`
`=x^2+2.x.2y+(2y)^2-z^2`
`=x^2+4xy+4y^2-z^2`
Áp dụng HĐT:`A^2-B^2=(A+B)(A-B)`
`(A+B)^2=A^2+2AB+B^2`
a) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\cdot\left[x\cdot\left(x-1\right)-\left(x^2-x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2-x-x^2+x-1\right)\)
\(=\left(x+1\right)\cdot\left(-1\right)\)
\(=-1\left(x+1\right)\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x+4\right)\left(x-4\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+\left(3x+12\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x+12x-12\)
\(=x^3-1-x^3-8+12x-12\)
\(=-21+12x\)
c) \(3x^2\left(x+1\right)\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=3x^2\left(x^2-1\right)+x^6-3x^4+3x^2-1-\left(x^6-1\right)\)
\(=3x^4-3x^2+x^6-3x^4+3x^2-1-x^6+1\)
\(=0\)
câu b bạn làm sai rồi í!