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a) \(\frac{3}{4}-1\frac{1}{2}+0,5\div\frac{5}{12}=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}=\frac{9}{20}\)
b) \(\left(-2\right)^2-1\frac{5}{27}\left(-\frac{3}{2}\right)^3=4-\frac{32}{27}.\frac{27}{8}=4-4=0\)
c) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
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\(\frac{3^4.2-3^6}{3^4.17+4.3^4}\)
\(=\frac{3^4.\left(2-9\right)}{3^4.\left(17+4\right)}\)
\(=\frac{-7}{21}=-\frac{1}{3}\)
HT
\(\frac{3^4.2-3^6}{3^4.17+\left(2-9\right)}\)
\(=\frac{3^4.\left(2+9\right)}{3^4.\left(17+4\right)}\)
\(=\frac{-7}{21}=-\frac{1}{3}\)
_HT_
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Xét hạng tổng quát:
\(\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{\sqrt{n}+\sqrt{n-1}}=\frac{\sqrt{n}-\sqrt{n-1}}{n-n+1}=\sqrt{n}-\sqrt{n-1}\)
Áp dụng vào bài, ta có:
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{n}-\sqrt{n-1}\right)\)
\(=\sqrt{n}-1\)
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\(\left(x+1\right).\left(3y-2\right)=-15\)
\(x,y\in Z\Rightarrow x+1;3y-2\inƯ\left(-15\right)=\left\{\pm1;\pm3\pm5\pm15\right\}\)
TA THẤY 3y-2 chia 3 dư 1 nên \(3y-2\in\left\{1;-5\right\}\)
ta có bảng sau
\(3y-2\) | \(1\) | \(-5\) |
\(x+1\) | \(-15\) | \(3\) |
\(x\) | \(-16\) | \(2\) |
\(y\) | \(1\) | \(-1\) |
bạn tự kết luận nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
đáp án :0,1/ 0,2/ 0,3/ 0,4/ 0,5/ 0,6/ 0,7/ 0,8 /0,9/ 0,10
\(\frac{2}{10}=0,2\)
\(\frac{3}{10}=0,3\)
\(\frac{4}{10}=0,1\)
\(\frac{5}{10}=0,5\)
\(\frac{6}{10}=0,6\)
\(\frac{7}{10}=0,7\)
\(\frac{8}{10}=0,8\) Mik ko viết đc tia nên mong bạn thông cảm
\(\frac{9}{10}=0,9\)
\(1=1,1\)