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HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)
b)
\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ = - 30\end{array}\)
26 tháng 4 2018
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
26 tháng 4 2018
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
TN
24 tháng 3 2017
Bài 1:
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+1986}\right)\)
Nhận xét: \(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Do đó: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+1986}\right)\)
\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot...\cdot\frac{1985\cdot1988}{1986\cdot1987}=\frac{1\cdot4\cdot1988}{1986\cdot3}=\frac{3976}{2979}\)
Bài 2:
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
\(\Rightarrow\frac{4\cdot4^5}{3\cdot3^5}\cdot\frac{6\cdot6^5}{2\cdot2^5}=2^x\)\(\Rightarrow\frac{4^6}{3^6}\cdot\frac{6^6}{2^6}=2^x\)
\(\Rightarrow\frac{\left(2^2\right)^6}{3^6}\cdot\frac{\left(2\cdot3\right)^6}{2^6}=2^x\)\(\Rightarrow\frac{2^{12}}{3^6}\cdot\frac{2^6\cdot3^6}{2^6}=2^x\)
\(\Rightarrow\frac{2^6\cdot3^6\cdot2^{12}}{2^6\cdot3^6}=2^x\)\(\Rightarrow2^{12}=2^x\Rightarrow x=12\)
XO
3 tháng 7 2019
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
= \(\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{-2}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{-4}{5}+\frac{4}{5}\right)+\frac{5}{6}-\frac{6}{7}\)
= \(\frac{5}{6}-\frac{6}{7}\)
= \(\frac{-1}{42}\)
4 tháng 7 2019
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{5}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{-2}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{-4}{5}+\frac{4}{5}\right)+\frac{5}{6}-\frac{6}{7}\)
\(=\frac{5}{6}-\frac{6}{7}\)
\(=\frac{-1}{42}\)
\(\frac{6}{5}+\left|\frac{1}{2}-1\right|\)
\(=\frac{6}{5}+\left|-\frac{1}{2}\right|\)
\(=\frac{6}{5}+\frac{1}{2}\)
\(=\frac{17}{10}\)
\(=\frac{5}{6}+\left|-\frac{1}{2}\right|\)
\(=\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{6}+\frac{3}{6}\)
\(=\frac{4}{3}\)