d) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)

\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)

\(=-3+\frac{3}{2}+1-5+\frac{6}{5}-\frac{3}{2}\)

\(=-7+\frac{6}{5}=-\frac{29}{5}\)

đề bài là gì vậy bạn

1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)

\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)

( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))

2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)

\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)

Vậy \(x=\frac{-92}{105}\)

b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)

Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)

b) \(\left(2x-1\right).\left(2x+3\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)

\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)

d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

=> \(x:\frac{1}{45}=\frac{1}{2}\)

=> \(x=\frac{1}{2}.\frac{1}{45}\)

=> \(x=\frac{1}{90}\)

Vậy \(x=\frac{1}{90}.\)

b) \(\left(2x-1\right).\left(2x+3\right)=0\)

=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)

Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.

Chúc bạn học tốt!

a) \(-\frac{4}{7}+\frac{\left(-5\right).\left(-39\right)}{13.25}+\frac{\left(-1\right).6}{42.\left(-5\right)}=-\frac{4}{7}+\frac{\left(-1\right).3}{1.5}+\frac{\left(-1\right).1}{7.\left(-5\right)}=-\frac{4}{7}+\frac{3}{5}+\frac{1}{35}\)

\(=-\frac{20}{35}+\frac{21}{35}+\frac{1}{35}=\frac{2}{35}\)

b) \(=\frac{2}{9}.\left[-\frac{4}{45}:\left(\frac{3}{15}-\frac{2}{15}\right)+1\frac{2}{3}\right]+\frac{5}{27}=\frac{2}{9}.\left[-\frac{4}{45}:\frac{1}{15}+1\frac{2}{3}\right]+\frac{5}{27}\)

\(=\frac{2}{9}.\left[\frac{\left(-4\right).15}{45.1}+1\frac{2}{3}\right]+\frac{5}{27}==\frac{2}{9}.\left[\frac{\left(-4\right).1}{3.1}+1\frac{2}{3}\right]+\frac{5}{27}\)

\(==\frac{2}{9}.\left[-\frac{4}{3}+\frac{5}{3}\right]+\frac{5}{27}=\frac{2.1}{9.3}+\frac{5}{27}=\frac{2}{27}+\frac{5}{27}=\frac{7}{27}\)

thanks nhiều nha

\(D=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

Làm tắt nha :

\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(D=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{98}{100}\)

\(D=\frac{1}{2}\left(\frac{98}{99}-\frac{98}{100}\right)\)

Tự tính nốt nha 

B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7  + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15

= 0 -0-0-0-0+7/9 +13/15

= 74/45

b, Nhóm các cặp trái dấu vào với nhau thì hết cuối cùng còn 13/15

c,\(\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...+\frac{1}{2}-\frac{1}{3}+1\)

= \(\frac{1}{6}+1\)= 7/6

b) -5/7 . 4/13 + -5/7 . 9/13 + -2/7

= -5/7 . (4/13 + 9/13) + -2/7

= - 5/7 + - 2/7

= -1

k nếu đúng, học tốt nha