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Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7
\(a,\frac{2}{3}\cdot x-\frac{4}{7}=\frac{1}{8}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{1}{8}+\frac{4}{7}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{7}{56}+\frac{32}{56}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{39}{56}\)
\(\Leftrightarrow x=\frac{39}{56}:\frac{2}{3}=\frac{39}{56}\cdot\frac{3}{2}=\frac{39\cdot3}{56\cdot2}=\frac{117}{112}\)
\(b,\frac{2}{7}-\frac{8}{9}\cdot x=\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{2}{7}-\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{6}{21}-\frac{14}{21}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{-8}{21}\)
\(\Leftrightarrow x=\frac{-8}{21}:\frac{8}{9}=\frac{-8}{21}\cdot\frac{9}{8}=\frac{-8\cdot9}{21\cdot8}=\frac{-1\cdot3}{7\cdot1}=\frac{-3}{7}\)
Làm nốt hai bài cuối đi nhé
Study well >_<
Mk k chép lại đề bài nha
a)\(\frac{2}{3}.x=\frac{1}{8}+\frac{4}{7}\)
\(\frac{2}{3}.x=\frac{7}{56}+\frac{32}{56}\)
\(\frac{2}{3}.x=\frac{39}{56}\)
\(x=\frac{39}{56}:\frac{2}{3}\)
\(x=\frac{39}{56}.\frac{3}{2}\)
\(x=\frac{117}{112}\)
Mk sợ sai lém!!!
Lê Minh Phương tham khảo bài mình nhé
\(a,\frac{9}{-7}< x>\frac{7}{2}\)
\(\Leftrightarrow\frac{-9}{7}< x>\frac{7}{2}\)
\(\Leftrightarrow\frac{-18}{14}< x>\frac{49}{14}\)
\(\Leftrightarrow-18< x>49\)
\(\Leftrightarrow x\in\left\{-17;-16;-15;...;50\right\}\)
Còn bài kia tương tự
\(a,\frac{9}{-7}< x< \frac{7}{2}\)
\(\Rightarrow\frac{9.2}{-7.2}< x< \frac{7.7}{2.7}\)
\(\Rightarrow\frac{-18}{14}< x< \frac{49}{14}\)
\(\text{vì}x\in Z\Rightarrow x=-\frac{14}{14};\frac{0}{14};\frac{14}{14};\frac{28}{14};\frac{42}{14}\)
\(\text{hay }x=\left\{-1;0;1;2;3\right\}\)
a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)
\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)
b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)
\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)
\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)
c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)
\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)
\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)
d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)
e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)
\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)
a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)
\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{303}{610}\)
\(\Rightarrow B=\frac{101}{610}\)
b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)
\(\Rightarrow C=\frac{408}{205}\)
c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)
a) \(\frac{1}{3}-\frac{-1}{6}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
b) \(2\frac{1}{3}+4\frac{1}{5}=\frac{7}{3}+\frac{21}{5}=\frac{98}{15}\)
c) \(\frac{4}{9}-\frac{13}{3}-\frac{4}{9}-\frac{10}{3}=\left(\frac{4}{9}-\frac{4}{9}\right)-\left(\frac{13}{3}+\frac{10}{3}\right)\)
\(=0-\frac{23}{3}=\frac{-23}{3}\)
d) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
\(x.\frac{3}{7}=\frac{2}{3}\)
\(x=\frac{2}{3}:\frac{3}{7}\)
\(x=\frac{14}{9}\)
\(24:\frac{-6}{11}=24.\frac{-11}{6}\)
\(=\frac{-264}{6}=-44\)