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

b. \(\left(2x-1\right)^2=\left(2x\right)^2-2.2x.1+1^2=4x^2-4x+1\)

c. \(\left(x+2\right)^3=x^3+3x^2.2+3x.2^2+2^3=x^3+6x^2+12x+8\)

d. \(\left(x-2\right)^3=x^3-3x^2.2+3x.2^2-2^3=x^3-6x^2+12x-8\)

e. \(\left(1-3x\right)^2=1^2-2.1.3x+\left(3x\right)^2=1-6x+9x^2\)

f. \(\left(2x+1\right)^3=\left(2x\right)^3+3\left(2x\right)^2.1+3.2x.1^2+1^3=8x^3+12x^2+6x+1\)

g. \(\left(3-x\right)^3=3^3-3.3^2x+3.3x^2-x^3=27-27x+9x^2-x^3\)

\(54^2+44^2-54\cdot88=54^2-2\cdot44\cdot54+44^2=\left(54-44\right)^2=10^2=100\)

1234^8×56743²⁷ = 1.21942e+153 nha bạn 

HT

Ta có

\(n-24⋮77\) => n-24 đồng thời chia hết cho 7 và 11

\(n-24⋮7\Rightarrow\left(n-3\right)-21⋮7\Rightarrow n-3⋮7\Rightarrow a=3\)

\(n-24⋮11\Rightarrow\left(n-2\right)-22⋮11\Rightarrow n-2⋮11\Rightarrow b=2\)

\(\Rightarrow a+b=3+2=5\)

TL

x=2

HT

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

\(\Rightarrow\left(x+3\right)+\left(3+x\right)^2=0\)

\(\Rightarrow\left(x+3\right)+\left(x+3\right)\)

\(\Rightarrow\left(x+3\right)\left(1+x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

bài 1

a) 2003.2005=(2004-1).(2004+1)=\(2004^2\)-1 (\(1\))

  \(2004^2\)(2)

từ 1 và 2 => 2003.2005<\(2004^2\)

b)....

Bài 2

2015.2017=(2016-1).(2016+1)=\(2016^2\)-1  (1)

\(2016^2\)(2)

từ 1 và 2 => \(2016^2\)> 2015.2017=> A>B

\(E=\left(x^2y+y^3\right)\left(x^3+xy^2\right)-xy\left(x^4+y^4\right)\)

\(=\left(x^5y+x^3y^3+x^3y^3+y^5\right)-xy^5-x^5y\)

\(=x^5y+x^3y^3+x^3y^3+xy^5-xy^5-x^5y\)

\(=2x^3y^3\)

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

\(=\left(1-x^3\right)\left(1+x^3\right)\)

\(=1+x^3-x^3-x^6\)

\(=1-x^6\)