tìm x: \(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\)
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\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{41}{10}\div\frac{9}{4}=\frac{41}{10}\times\frac{4}{9}=\frac{82}{45}\Rightarrow\frac{82}{45}=\frac{x}{7,3}\)
Đến đấy thôi
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a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
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\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
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\(b)\) Ta có: \(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45\text{ }}\)
\(\Leftrightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Leftrightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Leftrightarrow x-\frac{37}{45}=1\)
\(\Leftrightarrow x=1+\frac{37}{45}\)
\(\Leftrightarrow x=\frac{82}{45}\)
Vậy \(x=\frac{82}{45}\)
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Xét vế phải
\(\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\frac{9}{1}\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+9\)
\(=\frac{1}{9}+\frac{2}{8}+\frac{3}{7}+\frac{4}{6}+\frac{5}{5}+\frac{6}{4}+\frac{7}{3}+\frac{8}{2}+\left(1+1+...+1\right)\)
\(=\left(1+\frac{1}{9}\right)+\left(1+\frac{2}{8}\right)+\left(1+\frac{3}{7}\right)+\left(1+\frac{4}{6}\right)+\left(1+\frac{5}{5}\right)+\left(1+\frac{6}{4}\right)+\left(1+\frac{7}{3}\right)+\left(1+\frac{8}{2}\right)+1\)\(=\frac{10}{9}+\frac{10}{8}+\frac{10}{7}+\frac{10}{6}+\frac{10}{5}+\frac{10}{4}+\frac{10}{3}+\frac{10}{2}+\frac{10}{10}\)
\(=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)
Thay vào bài ,ta được:
\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\times x=10\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)\)\(\Rightarrow x=10\)
Vậy x=10
chúc bạn học tốt
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\(\frac{15}{41}+\frac{-138}{41}< x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\Leftrightarrow\frac{-123}{41}< x< \frac{1.3+1.2+1}{6}\)
\(\Leftrightarrow-3< x< 1\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
\(\frac{x}{5}=\frac{15}{2}-\frac{51}{10}\)
\(\frac{x}{5}=\frac{15.5-51}{10}\)
\(\frac{x}{5}=\frac{24}{10}\)
\(\frac{x}{5}=\frac{12}{5}\)
\(x=12\)
\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\)
\(\Rightarrow x.\frac{9}{4}=\frac{41}{10}.7,3\)
\(\Rightarrow x.\frac{9}{4}=\frac{2993}{100}\)
\(\Rightarrow x=\frac{2993}{100}:\frac{9}{4}\)
\(\Rightarrow x=\frac{2993}{225}\)
Vậy \(x=\frac{2993}{225}.\)
Chúc bạn học tốt!
\(\frac{\frac{41}{10}}{\frac{9}{4}}=\frac{x}{7,3}\\ \Rightarrow\frac{9}{4}x=\frac{41}{10}\cdot7,3\left(\text{tính chất của tỉ lệ thức}\right)\\ x=\frac{\frac{41}{10}\cdot7,3}{\frac{9}{4}}=\frac{2993}{225}\)
Vậy \(x=\frac{2993}{225}\)