A

\(\frac{3}{5}\div\frac{4}{5}+\frac{1}{2}\times\frac{2}{3}\)

\(=\frac{3}{4}+\frac{1}{3}\)

\(=\frac{13}{12}\)

B

\(\frac{5}{4}\times x=\frac{3}{8}+\frac{1}{4}\)

 \(\frac{5}{4}\times x =\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{5}{4}\)

\(x=\frac{4}{8}=\frac{1}{2}\)

a) chắc bạn cũng biết 

b) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{2005.\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005.1-1}{2004+2005.2006}=\frac{2005.2006+2004}{2004+2005.2006}=1\)

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

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

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

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

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

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

\(A=x\left(x+4\right)-6\left(x-1\right)\left(x+1\right)+\left(2x-1\right)^2\)

\(A=x^2+4x-6\left(x^2-1\right)+\left(4x^2-4x+1\right)\)

\(A=x^2+4x-6x^2+6+4x^2-4x+1\)

\(A=-x^2+7\)

Để A có giá trị bằng 3 thì :

\(-x^2+7=3\)

\(-x^2=-4\)

\(x^2=4\)

\(x\in\left\{\pm2\right\}\)

Vậy..........

giup minh voi cac ban