\(\left(x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-4\right)=0\)

\(\left(x-4\right)\left(x+3+x+5\right)=0\)

\(\left(x-4\right)\left(2x+8\right)=0\)

\(\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\\2x+8=0\Leftrightarrow x=-4\end{matrix}\right.\)

\(102-\left[8^2-48\right].0,5.\left(2^2.10+8\right):2^5\)

\(=102-\left[64-48\right].0,5.\left(40+8\right):32\)

\(=102-16.0,5.48:32\)

\(=102-\left(8.\dfrac{3}{2}\right)\)

\(=102-12\)

\(=90\)

cứu mình với

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{33}{99}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}\\ =\dfrac{32}{99}\)

Câu này bạn hỏi rồi mà dương

Cho bn í hỏi lại

= \(\dfrac{5}{2}\)

bạn THAM KHẢO thôi nhé!

Kq = 5/2 nha bạn câu này đã có người hỏi rồi bạn tự tham khảo nhé 

Hỏi rồi àm sao hỏi lại vậy

\(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right):\left(x-\dfrac{13}{7}\right)< 0\)

\(TH1:\left\{{}\begin{matrix}x-\dfrac{1}{5}>0\\x-\dfrac{13}{7}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{13}{7}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{5}< x< \dfrac{13}{7}\)

\(TH2:\left\{{}\begin{matrix}x-\dfrac{1}{5}< 0\\x-\dfrac{13}{7}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\) (vô lý nên loại)

Vậy \(\dfrac{1}{5}< x< \dfrac{13}{7}\) thỏa mãn đề bài

\(0,3\left(5\right)=\dfrac{3}{2}\)

\(0,3\left(5\right)=3,2.\dfrac{1}{9}=\dfrac{32}{10}.\dfrac{1}{9}=\dfrac{16}{45}\)