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Ta nhận thấy mẫu của biểu thức trên là:
x26+x24+x22+...+x2+1=(x26+x22+...+x2)+(x24+x20+...+x4+1)
=x2(x24+x20+...+x16+...+1)+(x24+x20+...+x4+1)
=(x24+x20+...+1)(x2+1)
Như vậy\(\frac{x^{24}+x^{20}+x^{16}+...+1}{\left(x^{24}+x^{20}+...+1\right)\left(x^2+1\right)}\)=\(\frac{1}{x^2+1}\)
\(B=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{x^{30}+x^{28}+x^{26}+...+x^2+1}\)
\(=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{\left(x^{30}+x^{26}+x^{22}+...+x^6+x^2\right)+\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)}\)
\(=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{x^2\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)+\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)}\)
\(=\frac{x^{28}+x^{24}+x^{20}+...+x^4+1}{\left(x^2+1\right)\left(x^{28}+x^{24}+x^{20}+...+x^4+1\right)}=\frac{1}{x^2+1}\)
a) Sửa đề \(\frac{-3}{x+1}=\frac{x+1}{-12}\)
<=> (x + 1)(x + 1) = (-12).(-3)
<=> (x + 1)2 = 36
<=> \(\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
b) \(\frac{x}{5}=-\frac{x+24}{3}\)
=> 3x = -(x + 24).5
<=> 3x = -5x - 120
<=> 8x = -120
<=> x = -15
Vậy x = -15
c) \(\frac{x+2}{x+1}=\frac{x-4}{x-2}\)
<=> \(\frac{x+2}{x+1}-1=\frac{x-4}{x-2}-1\)
<=> \(\frac{1}{x+1}=\frac{-2}{x-2}\)
<=> (x - 2).1 = -2(x + 1)
<=> x - 2 = -2x - 2
<=> 3x = 0
<=> x = 0
Vậy x = 0
d) \(\frac{x+4}{y+7}=\frac{4}{7}\)
<=> \(\frac{x+4}{4}=\frac{y+7}{7}=\frac{x+4+y+7}{4+7}=\frac{x+y+11}{11}=\frac{22+11}{11}=3\)(dãy tỉ số bằng nhau)
<=> \(\hept{\begin{cases}\frac{x+4}{4}=3\\\frac{y+7}{7}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x+4=12\\y+7=21\end{cases}}\Leftrightarrow\hept{\begin{cases}x=8\\y=14\end{cases}}\)
a ) \(-\frac{3}{x+1}=\frac{x+1}{-12}\)
\(\Leftrightarrow\)\(\left(x+1\right).\left(x+1\right)=-3.\left(-12\right)\)
\(\Leftrightarrow\)\(\left(x+1\right)^2=36\)
\(\Leftrightarrow\)\(\left(x+1\right)^2=\pm6\)
\(\Rightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
b ) \(\frac{x}{5}=\frac{x+24}{3}\)
\(\Leftrightarrow\)\(3x=\left(x+24\right).5\)
\(\Leftrightarrow\)\(3x=5x+120\)
\(\Leftrightarrow\)\(-2x=120\)
\(\Leftrightarrow\)\(x=-60\)
d ) \(\frac{x+4}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x+4}{4}=\frac{7+y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x+4}{4}=\frac{7+y}{7}=\frac{\left(x+y\right)+\left(4+7\right)}{4+7}=\frac{22+11}{11}=\frac{33}{11}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x+4}{4}=3\\\frac{7+y}{7}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+4=12\\7+y=21\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=8\\y=14\end{cases}}\)
\(\frac{149-x}{26}+\frac{171-x}{24}+\frac{189-x}{22}+\frac{203-x}{20}=10\)
\(\left(\frac{149-x}{26}-1\right)+\left(\frac{171-x}{24}-2\right)+\left(\frac{189-x}{22}-3\right)+\left(\frac{203-x}{20}-4\right)=0\)
\(\frac{123-x}{26}+\frac{123-x}{24}+\frac{123-x}{22}+\frac{123-x}{20}=0\)
\(\left(123-x\right)\left(\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\right)=0\)
\(123-x=0\left(vì\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\ne0\right)\)
\(\Rightarrow x=123\)
\(B=\frac{1+x^2+x^4+...+x^{26}}{1+x^4+x^8+...+x^{24}}\)
\(=\frac{\frac{\left(x^2-1\right)\left(1+x^2+x^4+...+x^{26}\right)}{x^2-1}}{\frac{\left(x^4-1\right)\left(1+x^4+x^8+...+x^{24}\right)}{x^4-1}}\)
\(=\frac{\frac{x^{28}-1}{x^2-1}}{\frac{x^{28}-1}{x^4-1}}=\frac{x^4-1}{x^2-1}=x^2+1\)