Bài 2 :

a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

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

\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)

\(A=\left(\frac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)

\(A=\frac{4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(x-2\right)^2}{4}\)

\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)

\(A=\frac{x-2}{x+2}\)

c) Thay x = 4 ta có :

\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)

Vậy.........

\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)

\(\left(5x-2\right)\left(25x^2+10x+4\right)\)

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

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

\(=125x^3-8\)

Bài 1 :

Ta có : \(\frac{x^2+x+1}{x^2+1}=0\)

=> \(\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{x^2+1}=0\)

Ta thấy \(\left\{{}\begin{matrix}\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\\x^2+1>0\end{matrix}\right.\)

=> \(\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{x^2+1}>0\)

Vậy phương trình vô nghiệm .

Bài 3 :

a, ĐKXĐ : \(\left\{{}\begin{matrix}m-2\ne0\\m\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}m\ne2\\m\ne0\end{matrix}\right.\)

Ta có : \(A=\frac{m+1}{m-2}-\frac{1}{m}\)

=> \(A=\frac{\left(m+1\right)m}{\left(m-2\right)m}-\frac{m-2}{m\left(m-2\right)}\)

=> \(A=\frac{m^2+m-m+2}{\left(m-2\right)m}=\frac{m^2+2}{m\left(m-2\right)}\)

Ta có : \(B=\frac{m+2}{m-2}+\frac{1}{m}\)

=> \(B=\frac{\left(m+2\right)m}{\left(m-2\right)m}+\frac{m-2}{m\left(m-2\right)}\)

=> \(B=\frac{m^2+2m+m-2}{\left(m-2\right)m}=\frac{m^2+3m-2}{m\left(m-2\right)}\)

c, Thay A = 1 ta được phương trình :\(\frac{m^2+2}{m\left(m-2\right)}=1\)

=> \(m^2+2=m\left(m-2\right)\)

=> \(-2m=2\)

=> \(m=-1\) ( TM )

Vậy m có giá trị bằng 1 khi A = 1 .

b, - Để A = B thì : \(\frac{m^2+2}{m\left(m-2\right)}=\frac{m^2+3m-2}{m\left(m-2\right)}\)

=> \(m^2+2=m^2+3m-2\)

=> \(3m=4\)

=> \(m=\frac{4}{3}\)

Vậy với A = B thì m có giá trị là 4/3 .

d, Ta có : A + B = 0 .

=> \(\frac{m^2+2}{m\left(m-2\right)}+\frac{m^2+3m-2}{m\left(m-2\right)}=0\)

=> \(2m^2+3m=0\)

=> \(m\left(2m+3\right)\)=0

=> \(\left[{}\begin{matrix}m=0\\m=-\frac{3}{2}\end{matrix}\right.\)

Vậy m = 0 hoăc m = -3/2 khi A + B = 0 .

Hack não

Bài 1 :

a) \(x^8+x+1\)

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

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

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

\(=\left(x^5+x^2\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=\left(x^5+x^2\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^6-x^5+x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^6-x^5+x^4-x^2+1\right)\left(x^2+x+1\right)\)

b) \(64x^4+y^4\)

\(=\left(8x^2\right)^2+\left(y^2\right)^2+2.8x^2.y^2-16x^2y^2\)

\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)