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a,(x+2y)3 =x3+3.x2.2y+3.x.(2y)2+(2y)3
= x3+6x2y+12xy2+8y3
b, phần b tương tự dấu thay đổi một tí
c, (5x+1)(5x+1)= (5x+1)2
=25x2+10x+1
a) \(\left(2x+1\right)^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)
\(=8x^3+12x^2+6x+1\)
b) \(\left(x-3\right)^3\)
\(=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
Bài 2:
a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)
b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)
c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)
a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36y^2x^2-8y^3\)
c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
4:
a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2
b: 8(7^2+1)(7^4+1)(7^8+1)
=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^16-1)<7^16-1
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
mik chỉ biết bài 5 thôi !
a) \(\left(3x^2-2y^3\right)^2\)
\(=\left(3x^2\right)^2-2\cdot3x^2\cdot2y^3+\left(2y^3\right)^2\)
\(=9x^4-12x^2y^3+4y^6\)
b) \(\left(-2x^2-3\right)^2\)
\(=\left(-2x^2\right)^2-2\cdot\left(-2x^2\right)\cdot3+3^2\)
\(=4x^4+12x^2+9\)
\(\left(5x+3yz\right)^2=25x^2+30xyz+9y^2z^2\\ \left(2x-3\right)^3=8x^3-36x^2+54x-27\)
\(\left(y^{2x}+3yz\right)^2=y^{4x}+6y^{2x+1}z+9y^2z^2\\ \left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\\ \left(y-5\right)\left(25+2y+y^2+3y\right)=\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)