1.(__3x_____+___1____) (9x^2-__3x____+1) = (__3x__)^3 + (__1____)^3

2.(x+__4_____) (__x\(^2\)____-___4x_____+16) = (_x___)^3 + (__4____)^3

hai câu cuối bạn có gõ sai chỗ 10x với 6x ko??? vì như thế thì ko ra đc HĐT

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 !

\(a,=3x\left(y-z\right)-y\left(y-z\right)=\left(3x-y\right)\left(y-z\right)\\ b,=x^3\left(x-1\right)+x\left(x-1\right)=x\left(x^2+1\right)\left(x-1\right)\\ c,=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\\ d,=\left(x-3\right)^2\\ e,=\left(x+2\right)^3\\ f,=\left(2x-x+y\right)\left(2x+x-y\right)=\left(x+y\right)\left(3x-y\right)\\ g,=\left(y+1\right)\left(5x-2\right)\\ h,=\left(x+2\right)^2\\ i,=x^2\left(x^2-2\right)\\ k,=3x\left(x-4y\right)\)

\(a,=\left(5x+1\right)^2-y^2=\left(5x-y+1\right)\left(5x+y+1\right)\\ b,Sửa:25\left(2x+1\right)^2-\left(3x+2\right)^2\\ =\left[5\left(2x+1\right)-3x-2\right]\left[5\left(2x+1\right)+3x+2\right]\\ =\left(10x+5-3x-2\right)\left(10x+5+3x+2\right)\\ \left(7x+3\right)\left(13x+7\right)\\ c,=2x^2-8x+3x-12=\left(x-4\right)\left(2x+3\right)\)

a: \(25x^2-\dfrac{10}{3}xy+\dfrac{1}{9}y^2=\left(5x-\dfrac{1}{3}y\right)^2\)

b: \(25x^2-15x+\dfrac{9}{4}=\left(5x-\dfrac{3}{2}\right)^2\)

c: \(\left(2x+\dfrac{1}{2}y\right)\left(4x^2-xy+\dfrac{1}{4}y^2\right)=8x^3+\dfrac{1}{8}y^3\)

d: \(\left(x^2-\dfrac{2}{3}\right)\left(x^4+\dfrac{2}{3}x^2+\dfrac{4}{9}\right)=x^6-\dfrac{8}{27}\)

a) \(3\in Z\)

b) \(-3\notin N\)

c) \(1\in N\)

d) \(N\subset Z\)

e) \(1;-2\in Z\)

bài này bn đăng rồi mà

a: \(3\in Z\)

b: \(-3\notin N\)

c: \(1\in N\)

d: \(N\subset Z\)

1)\(64x^2+144x+81=\left(8x+9\right)^2\)

2)\(16y^2+\frac{1}{64}+y=\left(4y+\frac{1}{8}\right)^2\)

3)\(9z^2+6z+1=\left(3z+1\right)^2\)

4)\(4y^2+12y+9=\left(2y+3\right)^2\)

5)\(y^2+10xy+25x^2=\left(y+5x\right)^2\)

6)\(x^2+6x+9=\left(x+3\right)^2\)