1/3 cộng x = 11/6 lưu ý dấu gạch chéo là dấu gạch phân số
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Bài làm:
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}.y=\frac{4}{3}\)
\(\Leftrightarrow y=\frac{4}{3}:\frac{10}{11}=\frac{4}{3}.\frac{11}{10}=\frac{22}{15}\)
Chú ý dấu \(\left(.\right)\)là dấu \(\left(\times\right)\)
Vậy \(y=\frac{22}{15}\)
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{10}{11}.y=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{11}.y=\frac{2}{3}\)
\(\Rightarrow y=\frac{2}{3}:\frac{5}{11}=\frac{22}{15}\)
LƯU Ý:các dấu chấm(.) là dấu nhân ^^.
Phép tính trên bằng: \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{x}{6x+9}\)
5/6+3/4 < 19/11 4/9 x 6/5 = 8/15
17/12-9/8 < 1/3 15/4:3/8 > 8
Ta có: \(\frac{5}{6}+\frac{3}{4}=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
Vì \(12>11\)
\(\Rightarrow\frac{19}{12}< \frac{19}{11}\)
\(\Rightarrow\frac{5}{6}+\frac{3}{4}< \frac{19}{11}\)
Ta có: \(\frac{17}{12}-\frac{9}{8}=\frac{34}{24}-\frac{27}{24}=\frac{7}{24}\)
Ta lại có: \(\frac{1}{3}=\frac{8}{24}\)
Vì \(8>7\)
\(\Rightarrow\frac{7}{24}< \frac{8}{24}\)
\(\Rightarrow\frac{17}{12}-\frac{9}{8}< \frac{1}{3}\)
Ta có: \(\frac{4}{9}\times\frac{6}{5}=\frac{24}{45}=\frac{8}{15}\)
Vì \(\frac{8}{15}=\frac{8}{15}\)
\(\Rightarrow\frac{4}{9}\times\frac{6}{5}=\frac{8}{15}\)
Ta có: \(\frac{15}{4}:\frac{3}{8}=\frac{15}{4}\times\frac{8}{3}=10\)
Vì \(10>8\)
\(\Rightarrow\frac{15}{4}:\frac{3}{8}>8\)
HOK TOT
3 x X - \(\frac{1}{3}\) = \(\frac{11}{6}\)
3 x X = \(\frac{11}{6}\)+ \(\frac{1}{3}\)
3 x X = \(\frac{13}{6}\)
X = \(\frac{13}{6}\):3
X=\(\frac{13}{18}\)
tk mk nha
3 x X = \(\frac{11}{6}\)+ \(\frac{1}{3}\)
3 x X = \(\frac{13}{6}\)
X = \(\frac{13}{6}\): 3
X = \(\frac{13}{18}\)
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{127}{128}\)
Ủng hộ mk nha ^_-
\(2\cdot A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)
Bạn thấy những ô phía trên khung đặt câu hỏi hay câu trả lời ko? Bạn vào cái ô công thức Toán ở cạnh ô Tex đó. Ở đó bạn chọn biểu tượng phân số là xong.
`x-10/3=7/18+3/5`
`=> x-10/3=89/90`
`=> x=89/90+10/3`
`=> x=389/90`
\(\dfrac{1}{3}+x=\dfrac{11}{6}\)
\(x=\dfrac{11}{6}-\dfrac{1}{3}\)
\(x=\dfrac{9}{6}\)
\(x=\dfrac{3}{2}\)
\(\dfrac{1}{3}+x=\dfrac{11}{6}\)
\(x=\dfrac{11}{6}-\dfrac{1}{3}\)
\(x=\dfrac{3}{2}\)