tính
\(\frac{13}{45}x\frac{15}{26}-\frac{13}{45}x\left(1\frac{1}{26}\right)+\frac{11}{26}x\frac{13}{45}\)
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\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
\(\frac{\left(\frac{518}{19}-\frac{342}{13}\right).\left(\frac{177}{236}+\frac{76}{236}-\frac{6}{236}\right)}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)
=>\(\frac{\left(\frac{6734}{247}-\frac{6498}{247}\right).\frac{247}{236}}{\left(\frac{3}{4}+x\right).\frac{27}{33}}=1\)
=>(3/4+x)*27/33=236/247*247/236=1
3/4+x=1:27/33=33/27
x=33/27-3/4=132/108-81/108
x=51/108
Vậy x=51/108
Ta có : \(C=\left(1+\frac{11}{13}\right).\left(1+\frac{11}{28}\right).\left(1+\frac{11}{45}\right)......\left(1+\frac{11}{220}\right)\)
\(=\frac{24}{13}.\frac{39}{28}.\frac{56}{45}......\frac{231}{220}\)
\(=\frac{2.12}{1.13}.\frac{3.13}{2.14}.\frac{4.14}{3.15}......\frac{11.21}{10.22}\)
\(=\frac{2.12.3.13.4.14.....11.21}{1.13.2.14.3.15.....10.22}\)
\(=\frac{\left(2.3.4.....11\right).\left(12.13.14.....21\right)}{\left(1.2.3.....10\right).\left(13.14.15.....22\right)}\)
\(=\frac{11.12}{1.22}=\frac{12}{1.2}=6\)
Vậy C = 6
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
a) 0,4:x=x:0,9
0,4.0,9=x2
0,36 =x2
\(\sqrt{0,36}\)=x
0,6 =x
= \(\frac{13}{45}x\left(\frac{15}{26}-1\frac{1}{26}+\frac{11}{26}\right)=\frac{13}{45}x\left(\frac{15}{26}-\frac{27}{26}+\frac{11}{26}\right)=\frac{13}{45}x\left(\frac{27}{26}-\frac{27}{26}\right)=\frac{13}{45}x0=0\)