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

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

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

a) \(x^6-y^6=\text{(x-y)(y+x)(y^2-xy+x^2)(y^2+xy+x^2)}\)                                                                                                            b)\(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

a)\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

b)\(x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2\)

2. Ta có : \(\hept{\begin{cases}a+b+c=2016\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2016}\end{cases}\Leftrightarrow}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow a+b=0\) hoặc \(b+c=0\) hoặc \(a+c=0\)

• Nếu a + b = 0 => c = 2016 (1)
• Nếu b + c = 0 => a = 2016 (2)
• Nếu a + c = 0 => b = 2016 (3)

Từ (1) , (2) , (3)  ta có điều phải chứng minh.