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

\(=x^2-1+x^3+2x^2+4x+2x^2+4x+8-x^3-x^2-2x\)

\(=\left(x^3-x^3\right)+\left(x^2+2x^2+2x^2-x^2\right)+\left(4x+4x-2x\right)+\left(-1+8\right)\)

\(=4x^2+6x+7\)

b) Thay vào ta được

\(A=4.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}+7=1+3+7=11\)

bài 1 : a +b , rút gọn và tính

(-a+b-c)-(a-b-c)= -a+b -c-a+b+c= -2a+2b= -2.1+2.-1=-2+-2 = -4

Ta có: \(B=1+2^2+2^4+.....+2^{18}\)

\(\Rightarrow2B=2+2^3+2^5+...+2^{19}\)

\(\Rightarrow2B-B=\left(2+2^3+2^5+....+2^{19}\right)-\left(1+2^2+2^4+...+2^{18}\right)\)

\(\Rightarrow B=2^{19}-1\)

Vậy rút gọn biểu thức \(B=1+2^2+2^4+...+2^{18}\) được \(2^{19}-1\)

B  =1+2^2+2^4+2^6+...+2^18

    =2^0+2^2+2^4+2^6+...+2^18

4B=2^2+2^4+2^6+2^8+...+2^20

4B-B=(2^2+2^4+2^6+2^8+...+2^20)-(2^0+2^2+2^4+2^6+...+2^18)

3B=2^2+2^4+2^6+2^8+...+2^20-2^0-2^2-2^4-2^6-...-2^18

3B=2^20-2^0

3B=2^20-1

B=(2^20-1)/3

\(A=100^2-99^2+98^2-97^2+....+2^2-1^2\)

\(=\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+....+\left(2-1\right).\left(2+1\right)\)

\(=1+2+....+97+98+99+100=\frac{100.\left(100+1\right)}{2}=5050\)

\(B=3\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)......\left(2^{64}+1\right)+1=\left(2^8-1\right).....\left(2^{64}+1\right)+1\)

Tiếp tục rút gọn như vậy,ta đc \(B=\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1+1=2^{128}\)

a, Với x khác 1 

\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)

b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

Vậy với x khác 1 thì bth A luôn nhận gtri âm