\(\left(\frac{13}{21}+\frac{17}{20}\right):\frac{176}{95}-\left(\frac{13}{20}-\frac{28}{21}\right):\frac{176}{96}\)
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1/Tìm x, biết :
a/ \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}+\frac{x}{\left(x+3\right)\left(x+34\right)}\)
b/ \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
1.Tìm x :
a,frac{1}{5.6}+frac{1}{6.7}+...+frac{1}{xleft(x+1right)}frac{13}{90}
b,frac{1}{1.4}+frac{1}{4.7}+frac{1}{7.10}+...+frac{1}{xleft(x+3right)}frac{49}{148}
c,frac{7}{left(x+3right)left(x+10right)}+frac{11}{left(x+10right)left(x+21right)}+frac{1}{left(x+21right)left(x+34right)}frac{x}{left(x+3right)left(x+34right)}
d,frac{3}{left(x-4right)left(x-7right)}+frac{6}{left(x-7right)left(x-13right)}+frac{15}{left(x-13right)left(x-28right)}-frac{1}{x-38}frac{-1}{20}
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1.Tìm x :
a,\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
b,\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
c,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}\)\(+\frac{1}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
d,\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}\)\(+\frac{15}{\left(x-13\right)\left(x-28\right)}\)\(-\frac{1}{x-38}=\frac{-1}{20}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
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\(\frac{1}{2}+\left(\frac{16}{21}+\frac{17}{13}\right)-\left(\frac{14}{13}-\frac{5}{21}\right)=?\)
Tính giá trị biểu thức1.Afrac{1}{5}+frac{3}{17}-frac{4}{3}+left(frac{4}{5}-frac{3}{17}+frac{1}{3}right)-frac{1}{7}+left[frac{-14}{30}right]2.Bleft(frac{5}{8}-frac{4}{12}+frac{3}{2}right)-left(frac{5}{8}+frac{9}{13}right)-left[frac{-3}{2}right]+frac{7}{-15}3.Cfrac{5}{18}+frac{8}{19}-frac{7}{21}+left(frac{-10}{36}+frac{11}{19}+frac{1}{3}right)-frac{5}{8}4.Dfrac{1}{9}-left[frac{-5}{23}right]-left(frac{-5}{23}+frac{1}{9}+frac{25}{7}right)+frac{50}{14}-frac{7}{30}5.Efrac{1}{13}+left(frac{-5}{18}-frac...
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Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
tìm x biết:
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}\)\(+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
B)\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{5}{2}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)
\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)
\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)
\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)
=> 5x - 20 = 2
=> 5x = 22
\(\Rightarrow x=\frac{22}{5}=4,4\)
Vậy, x = 4,4
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\(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-\frac{20}{15\cdot17}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
Sửa đề : \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\)\(x+1=18\)
\(\Leftrightarrow\)\(x=18-1\)
\(\Leftrightarrow\)\(x=17\)
Vậy \(x=17\)
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\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+10.\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x+\frac{40}{55}=\frac{3}{11}\)
\(\Leftrightarrow\)\(x=\frac{3}{11}-\frac{40}{55}\)
\(\Leftrightarrow\)\(x=\frac{-5}{11}\)
Vậy \(x=\frac{-5}{11}\)
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\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{3}-\frac{2}{x+1}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}=\frac{1}{9}\)
=> 2.9 = x + 1
=> x + 1 = 18
=> x = 17
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\(\frac{\frac{3}{67}\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{\left(29^3:100-29^3:0,47\right)}\)
\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
ghi sai để nên mik ko giải dc