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Giúp em bài toán này với :
Bài 3: Tìm x :
b) X x \(\frac{1}{2}\)+ \(\frac{3}{2}\)x X = \(\frac{4}{5}\)
\(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(\Rightarrow x.1=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Rightarrow x+\frac{3}{4}=\frac{10}{9}:\frac{5}{7}=\frac{10}{9}\times\frac{7}{5}=\frac{14}{9}\)
\(\Rightarrow x=\frac{14}{9}-\frac{3}{4}=\frac{56-27}{36}=\frac{29}{36}\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Leftrightarrow x+\frac{3}{4}=\frac{14}{9}\)
\(\Rightarrow x=\frac{29}{36}\)
P/s tham khảo nha
\(\frac{2}{3\times5}\times a+\frac{2}{5\times7}\times a+...+\frac{2}{13\times15}\times a=\frac{28}{15}\)
=> \(\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}\right)\times x=\frac{28}{15}\)
=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\times x=\frac{28}{15}\)
=> \(\left(\frac{1}{3}-\frac{1}{15}\right)\times x=\frac{28}{15}\)
=> \(\frac{4}{15}\times x=\frac{28}{15}\)
=> \(x=\frac{28}{15}:\frac{4}{15}\)
-> \(x=7\)
\(\frac{2}{3\times5}\times a+\frac{2}{5\times7}\times a+...+\frac{2}{13\times15}\times a=\frac{28}{15}\)
\(a\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}\right)=\frac{28}{15}\)
\(a\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{28}{15}\)
\(a\times\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{28}{15}\)
\(a\times\frac{4}{15}=\frac{28}{15}\)
\(a=\frac{28}{15}:\frac{4}{15}\)
\(a=\frac{28}{15}\times\frac{25}{4}\)
\(a=\frac{28}{4}=7\)
Bài làm
\(\frac{2}{3\times x}=\frac{3}{9}\)
\(\Leftrightarrow3\times x\times3=2\times9\)
\(\Leftrightarrow9\times x=18\)
\(\Leftrightarrow x=2\)
Đáp số \(x=2\)
\(x.\left(\frac{1}{6}.\frac{72}{10}+\frac{13}{10}+\frac{1}{2}\right)+15=19.75\)
\(\Leftrightarrow x.\left(\frac{6}{5}+\frac{13}{10}+\frac{1}{2}\right)=4,75\)
\(\Leftrightarrow x.3=4,75\) \(\Rightarrow x=1,583\)
Ủa mà có bài thì tự đi mà làm bài này có khó lắm đâu
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)
#)Giải :
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{10}\)
\(A=\frac{1}{10}\)
\(x\times\frac{1}{2}+\frac{3}{2}\times x=\frac{4}{5}\)
\(x\times\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(x\times2=\frac{4}{5}\)
\(x=\frac{4}{5}\div2\)
\(x=\frac{4}{10}\)