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\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
a) gt \(\Leftrightarrow\) s-\(10\times\left(\frac{2}{11\times13}+\frac{2}{13\times15}+...+\frac{2}{53\times55}\right)=\frac{3}{11}\)
\(\Leftrightarrow s-10\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow S-10\times\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow S=1\)
câu b hình như sai đề
Phải là \(\frac{1}{36}\) chứ ko phải \(\frac{1}{39}\)
\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{40}{55}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{40}{55}\)
\(x=\frac{55}{55}=1\)
nha.
A. = 1/2-1/3+1/3-1/4+1/4-1/5...+1/101-1/102=1/2-1/102=25/51.
B. =1/5-1/10+1/10-1/15+...+1/115-1/120=1/5-1/120=23/120.
C. = 1/5-1/7+1/7-1/9+1/9-1/11+...+1/997-1/999=1/5-1/999=994/4995.
Minh kiem tra bang may tinh roi do.
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{101\times102}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}\)
\(=1-\frac{1}{102}\)
\(=\frac{101}{102}\)
thế này :
= \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{11.13}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{13}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
= \(\frac{1}{2}.\frac{10}{39}\)
= \(\frac{5}{39}\)
Vậy kq = \(\frac{5}{39}\)
A= 7/5*7 + 7/7*9 + ... + 7/53*55
A= 7/2*( 2/5*7 + 2/7*9 + ... + 2/53*55 )
A= 7/2*( 7-5/5*7 + 9-7/7*9 + ... + 55-53/53*55 )
A= 7/2*( 1/5-1/7 + 1/7-1/9 + ... + 1/53-1/55 )
A= 7/2*( 1/5-1/55 )
A= 7/2*2/11
A= 7/11
A= 7/11 > 1/2
Nên: A > 1/2
B= 1/3 + 1/15 + 1/35 + ... + 1/99
B= 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/9*11
B= 2*( 2/1*3 + 2/3*5 + 1/5*7 + ... + 2/9*11 )
B= 2*( 3-1/1*3 + 5-3/3*5 + 7-5/5*7 + ... + 11-9/9*11 )
B= 2*( 1/1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/9-1/11 )
B= 2*( 1/1-1/11 )
B= 2*10/11
B= 20/11
B= 20/11 < 1/2
Nên: B < 1/2
A= 7/5*7 + 7/7*9 + ... + 7/53*55
A= 7/2*( 2/5*7 + 2/7*9 + ... + 2/53*55 )
A= 7/2*( 7-5/5*7 + 9-7/7*9 + ... + 55-53/53*55 )
A= 7/2*( 1/5-1/7 + 1/7-1/9 + ... + 1/53-1/55 )
A= 7/2*( 1/5-1/55 )
A= 7/2*2/11
A= 7/11
A= 7/11 > 1/2
Nên: A > 1/2
B= 1/3 + 1/15 + 1/35 + ... + 1/99
B= 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/9*11
B= 2*( 2/1*3 + 2/3*5 + 1/5*7 + ... + 2/9*11 )
B= 2*( 3-1/1*3 + 5-3/3*5 + 7-5/5*7 + ... + 11-9/9*11 )
B= 2*( 1/1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/9-1/11 )
B= 2*( 1/1-1/11 )
B= 2*10/11
B= 20/11
B= 20/11 < 1/2
Nên: B < 1/2
\(\frac{5}{2}+\frac{4}{11}+\frac{1}{11}+\frac{1}{30}+\frac{13}{60}=\frac{5}{2}+\frac{1}{30}+\frac{13}{60}+\frac{4}{11}+\frac{1}{11}\)
\(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}+\frac{5}{11}=\frac{33}{12}+\frac{5}{11}\)
\(\frac{363}{132}+\frac{60}{132}=\frac{423}{132}=\frac{36}{11}\)
