\(\frac{x^{30}+x^{28}+x^{26}+x^{24}+...+x^4+x^2+1}{x^{28}+x^{24}+x^{20}+...+x^8+x^4+1}=\frac{\left(x^{30}+x^{26}+x^{22}+...+x^2\right)+\left(x^{28}+x^{24}+...+x^4+1\right)}{x^{28}+x^{24}+x^{20}+...+x^4+1}\)

\(=\frac{x^2\left(x^{28}+x^{24}+...+x^4+1\right)+\left(x^{28}+x^{24}+...+x^4+1\right)}{x^{28}+x^{24}+...+x^4+1}\)

\(=\frac{\left(x^2+1\right)\left(x^{28}+x^{24}+...+x^4+1\right)}{x^{28}+x^{24}+...+x^4+1}\)

\(=x^2+1\)

\(B=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{x^{30}+x^{28}+x^{26}+...+x^2+1}\)

\(=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{\left(x^{30}+x^{26}+x^{22}+...+x^6+x^2\right)+\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)}\)

\(=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{x^2\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)+\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)}\)

\(=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{\left(x^2+1\right)\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)}=\frac{1}{x^2+1}\)

tự làm mẹ đê hỏi lắm

Xét \(x\ne1\)

Đặt \(y=x^4\).\(M=x^{28}+x^{24}+...+x^4+1\)

\(M=y^7+y^6+...+y^2+y+1\)\(\Rightarrow Ay=y^8+y^7+...+y^2+y\)

\(\Rightarrow M\left(y-1\right)=y^8-1\Rightarrow M=\frac{y^8-1}{y-1}=\frac{x^{32}-1}{x^4-1}\)

Tương tự \(N=x^{30}+x^{28}+...+x^2+1=\frac{\left(x^2\right)^{16}-1}{x-1}=\frac{x^{32}-1}{x-1}\)

\(A=\frac{M}{N}=\frac{\frac{x^{32}-1}{x^4-1}}{\frac{x^{32}-1}{x^2-1}}=\frac{x^2-1}{x^4-1}=\frac{1}{x^2+1}\)

Thay số vô tính ra A.

a) Ta có: \(\left(4x-1\right)^2=\left(1-4x\right)^2\)

\(\Leftrightarrow\left(4x-1\right)^2-\left(1-4x\right)^2=0\)

\(\Leftrightarrow\left(4x-1-1+4x\right)\left(4x-1+1-4x\right)=0\)

\(\Leftrightarrow0\cdot x=0\)(luôn đúng)

Vậy: \(x\in R\)

b) Ta có: \(\dfrac{x-100}{24}+\dfrac{x-98}{26}+\dfrac{x-96}{28}=3\)

\(\Leftrightarrow\dfrac{x-100}{24}-1+\dfrac{x-98}{26}-1+\dfrac{x-96}{28}-1=0\)

\(\Leftrightarrow\dfrac{x-124}{24}+\dfrac{x-124}{26}+\dfrac{x-124}{28}=0\)

\(\Leftrightarrow\left(x-124\right)\cdot\left(\dfrac{1}{24}+\dfrac{1}{26}+\dfrac{1}{28}\right)=0\)

mà \(\dfrac{1}{24}+\dfrac{1}{16}+\dfrac{1}{28}>0\)

nên x-124=0

hay x=124

Vậy: x=124

Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?