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) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\frac{3^4.2-3^6}{3^4.17+4.3^4}\)

\(=\frac{3^4.\left(2-9\right)}{3^4.\left(17+4\right)}\)

\(=\frac{-7}{21}=-\frac{1}{3}\)

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_

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:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(=\sqrt{n}-1\)

\(\left(x+1\right).\left(3y-2\right)=-15\)

\(x,y\in Z\Rightarrow x+1;3y-2\inƯ\left(-15\right)=\left\{\pm1;\pm3\pm5\pm15\right\}\)

TA THẤY 3y-2 chia 3 dư 1 nên \(3y-2\in\left\{1;-5\right\}\)

ta có bảng sau

bạn tự kết luận nhé

5cm nha

5cm  k cho m