\(\frac{4}{l2}\)= \(\frac{.....}{6}\) = \(\frac{l}{....}\)
giúp với hướng dữn mik lm với =(
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\(\frac{4\frac{1}{2}.5\frac{2}{3}}{6\frac{3}{4}}=\frac{\frac{9}{2}.\frac{17}{3}}{\frac{27}{4}}=\frac{\frac{153}{6}}{\frac{27}{4}}=\frac{153}{6}.\frac{4}{27}=\frac{34}{9}\)
\(\frac{4\frac{1}{2}\times5\frac{2}{3}}{6\frac{3}{4}}\)
\(\frac{\frac{4\times2+1}{2}+\frac{5\times3+2}{3}}{\frac{6\times4+3}{4}}\)
\(=\frac{\frac{9}{2}+\frac{17}{3}}{\frac{27}{4}}\)
\(=\frac{\frac{27}{6}+\frac{34}{6}}{\frac{27}{4}}\)
\(=\frac{61}{6}\div\frac{27}{4}=\frac{61}{6}\times\frac{4}{27}=\frac{244}{162}=\frac{122}{81}\)
Em nhân từng phân số với \(\frac{1}{7}\)
\(\frac{1}{7}P=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}+\frac{15}{28.43}+\frac{13}{43.56}\)
\(\frac{1}{7}P=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{56}\)
\(\frac{1}{7}P=\frac{1}{2}-\frac{1}{56}\)
\(\frac{1}{7}P=\frac{27}{56}\)
\(P=\frac{27}{56}:\frac{1}{7}\)
\(P=\frac{27}{8}>3\)
Vậy P >3
( ko hiểu chỗ nào thì hỏi nhá )
Rút gọn mỗi số hãng của số ta được :
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
Vậy C = 100/101
\(C=\frac{4}{1.2.3}+\frac{8}{3.4.5}+\frac{12}{5.6.7}+...+\frac{200}{99.100.101}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{101}{101}-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
a) \(\frac{3}{-4}=\frac{-3}{4};\frac{-1}{-4}=\frac{1}{4}\)
Vì - 3 < 1 nên \(\frac{-3}{4}< \frac{1}{4}\)
hay \(\frac{3}{-4}< \frac{-1}{-4}\)
Quy đồng mẫu ta được:
15/17=15.27/17.27=405/459
25/27=25.17/27.27=425/459
⇒405/459<425/459⇒15/17<25/27
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{7}\right]\)
\(Q=\left(\frac{2}{25}-\frac{126}{125}\right):\frac{4}{7}:\frac{-68}{9}\)
\(Q=\frac{-116}{125}:\frac{4}{7}:\frac{-68}{9}\)
\(Q=\frac{-203}{125}:\frac{-68}{9}\)
\(Q=\frac{1827}{8500}\)
a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)
b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)
TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)
Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)
a,\(\frac{3}{4}-x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)
TH1:\(x+\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
TH2:\(x+\frac{2}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
4/12=2/6=1/3