1./ \(x+y=3\Rightarrow\left(x+y\right)^3=27\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\Rightarrow x^3+y^3+3\cdot2\cdot3=27.\)

\(\Rightarrow x^3+y^3=9\)

2./ \(\left(x+3\right)\left(x^2-3x+3^2\right)-x^3-2x-4=0\)

\(\Leftrightarrow x^3+27-x^3-2x-4=0\Leftrightarrow2x=23\Leftrightarrow x=\frac{23}{2}\)

1/ \(x+y=3\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x^2+2xy+y^2=9\)

\(\Rightarrow x^2+4+y^2=9\)

\(\Rightarrow x^2+y^2=5\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.1=3\)

Ta có số a chia 7 dư 3 , tức là \(a=7k+3\left(k\in N\right)\)

\(\Rightarrow a^2=\left(7k+3\right)^2=\left(7k\right)^2+3^2+2.7.3k=7\left(7k^2+6k+1\right)+2=7Q+2\) 

Vậy a² chia 7 dư 2

ta có a:7 dư 3 nên a sẽ có dạng tổng quát là a=7k+3 \(\left(k\in N\right)\)

\(\Rightarrow\)a²=(7k+3)²=(7k)²+2.7k.3+7+2=7(7k²+6k+1)+2 ( có dạng B.Q+R)

vậy nên a²:7 dư 2

x + 40 =100

x = 100 - 40

x = 60

x+40=100

x=100-40

x=60

a) \(\left(x+1\right)^4-\left(x-1\right)^4=\left[\left(x+1\right)^2\right]^2-\left[\left(x-1\right)^2\right]^2\)

\(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right].\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)

\(=\left(x+1-x+1\right)\left(x+1+x-1\right)\left(x^2+2x+1+x^2-2x+1\right)\)

\(=2.2x.\left(2x^2+2\right)=8x\left(x^2+1\right)\)

b) \(\left(x^2-25\right)^2-4\left(x+5\right)^2=\left[\left(x-5\right)\left(x+5\right)\right]^2-4\left(x+5\right)^2\)

\(=\left(x+5\right)^2\left[\left(x-5\right)^2-4\right]=\left(x+5\right)^2\left(x^2-10x+25-4\right)=\left(x+5\right)^2\left(x^2-10+21\right)\)

\(=\left(x+5\right)^2\left(x-3\right)\left(x-7\right)\)