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\(a,2,5\cdot x=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}:2,5\)
\(\Rightarrow x=\frac{11}{15}:\frac{25}{10}\)
\(\Rightarrow x=\frac{11}{15}\cdot\frac{10}{25}\)
\(\Rightarrow x=\frac{11}{3}\cdot\frac{2}{25}=\frac{22}{75}\)
\(b,x-15\%\cdot x=\frac{1}{3}\)
\(\Rightarrow x-\frac{15}{100}\cdot x=\frac{1}{3}\)
\(\Rightarrow x-\frac{3}{20}\cdot x=\frac{1}{3}\)
\(\Rightarrow\frac{20x}{20}-\frac{3x}{20}=\frac{1}{3}\)
\(\Rightarrow\frac{17x}{20}=\frac{1}{3}\)
\(\Rightarrow17x\cdot3=20\)
\(\Rightarrow17x=\frac{20}{3}\)
\(\Rightarrow x=\frac{20}{3}:17=\frac{20}{3}\cdot\frac{1}{17}=\frac{20}{51}\)
Câu c mk làm sau :v
c,\(\left[\frac{3x}{7}+1\right]:\left[-4\frac{1}{7}\right]=\frac{-3}{28}\)
\(\Rightarrow\left[\frac{3x}{7}+1\right]:\left[-\frac{29}{7}\right]=\frac{-3}{28}\)
\(\Rightarrow\frac{3x}{7}+1=-\frac{3}{28}\cdot-\frac{29}{7}=\frac{87}{196}\)
\(\Rightarrow\frac{3x}{7}=\frac{87}{196}-1\)
Tìm nốt :v
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
=2/3.5/6.9/10.14/15.20/21.27/28
=5/9.21/15.45/49
=7/9.45/49
=5/7
hơi dài dòng nhưng mong bạn tích đúng cho mk nha
\(A=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot...\cdot\frac{1325}{1326}\)
\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot...\cdot\frac{2650}{2652}\)
\(=\frac{5\cdot8}{6\cdot7}\cdot\frac{6\cdot9}{7\cdot8}\cdot\frac{7\cdot10}{8\cdot9}\cdot...\cdot\frac{50\cdot53}{51\cdot52}\)
\(=\frac{5\cdot53}{7\cdot51}=\frac{265}{357}\)
Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)thỏa mãn đề.
\(\frac{-1}{21}+\frac{-1}{28}=\frac{-28}{588}+\frac{-21}{588}=\frac{-49}{588}=\frac{-1}{12}\)
Mình chưa học , mình mới lớp 5