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Ta có : \(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}=\frac{x-7}{59}\)
\(\Leftrightarrow\left(\frac{x-1}{65}-1\right)+\left(\frac{x-3}{63}-1\right)=\left(\frac{x-5}{61}-1\right)+\left(\frac{x-7}{59}-1\right)\)
\(\Leftrightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Leftrightarrow\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}=0\)
\(\Leftrightarrow\left(x-66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Mà ; \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)\ne0\)
Nên x - 66 = 0
=> x = 66
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\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{16}\right)+\left(1+\frac{1}{32}\right)+\left(1+\frac{1}{64}\right)-7\)
\(S=\left(1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-7\)
\(S=6+\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+....+\left(\frac{1}{32}-\frac{1}{64}\right)\right]-7\)
\(S=6+\left(1-\frac{1}{64}\right)-7\)
\(S=6+\frac{63}{64}-7\)
\(S=\frac{447}{64}-7=-\frac{1}{64}\)
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\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-1\)
\(S+1=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(2\left(S+1\right)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(2\left(S+1\right)-\left(S+1\right)=S+1=1-\frac{1}{2^6}=\frac{63}{64}\)
\(S=\frac{63}{64}-1\)
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a) Ta có:
\(\begin{array}{l}\frac{{123}}{7} = \frac{{123.4}}{{7.4}} = \frac{{492}}{{28}}\\17,75 = \frac{{1775}}{{100}} = \frac{{71}}{4} = \frac{{71.7}}{{4.7}} = \frac{{497}}{{28}}\end{array}\)
Vì 492 < 497 nên \(\frac{{492}}{{28}} < \frac{{497}}{{28}}\) hay \(\frac{{123}}{7} < 17,75\)
b) Ta có:
\(\begin{array}{l} - \frac{{65}}{9} = \frac{{( - 65).8}}{{9.8}} = \frac{{ - 520}}{{72}}\\ - 7,125 = \frac{{ - 7125}}{{1000}} = \frac{{ - 57}}{8} = \frac{{ - 57.9}}{{8.9}} = \frac{{ - 513}}{{72}}\end{array}\)
Vì 520 > 513 nên -520 < -513. Do đó, \(\frac{{ - 520}}{{72}} < \frac{{ - 513}}{{72}}\) hay \( - \frac{{65}}{9}\) < -7,125
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Ta có : \(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^{10}.26}=\frac{78}{26}=3\)
Ta có : \(B=\frac{4^6.9^5.6^9.120}{8^4.3^{12}+6^{11}}=\frac{4^6.9^5.120.6^9}{2^{12}.3^{12}+6^{11}}=\frac{4^6.9^5.6^9.120}{6^{12}+6^{11}}=\frac{6^9.9^5.4^6.120}{6^{11}\left(6+1\right)}\)
\(=\frac{9^5.4^6.120}{6^2.7}=\frac{\left(9.4\right)^5.480}{36.7}=\frac{36^4.480}{7}\)
( câu b đến đây hết rút được gồi / có chi sai thông cảm hen )