Bài 2: Cộng, trừ số hữu tỉ

=76,17

Bạn chỉ cần bấm máy tính rồi điền câu trả lời thôi, chắc không cần trình bày kĩ càng ra đâu. Tick cho mình nha

=86,17

76,17

a. Sắp xếp:

\(-\dfrac{7}{9};\dfrac{-40}{-50};\dfrac{27}{33};\dfrac{-6}{-4}\)

b. Sắp xếp:

\(-\dfrac{14}{37};-\dfrac{14}{33};0;\dfrac{18}{19};\dfrac{17}{20};\dfrac{4}{3}\)

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a)

Vì x là số hữu tỉ nên tập hợp \(-1< x< 0\) là vô số.

Ở đây, ta chỉ lấy 5 ví dụ cụ thể (theo đề bài):

\(-1< -\dfrac{3}{4};-\dfrac{1}{2};-\dfrac{7}{10};-\dfrac{12}{17};-\dfrac{3}{5};...< 0\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{2};-\dfrac{7}{10};-\dfrac{12}{17};-\dfrac{3}{5};...\right\}\)

b)

Có:

\(-\dfrac{4}{5}< x< -\dfrac{1}{10}\)

Hay \(-0,8< x< -0,1\)

Vì x là số hữu tỉ nên tập hợp \(-0,8< x< -0,1\)
là vô số.

Ở đây, ta chỉ lấy 5 ví dụ cụ thể (theo đề bài):

\(-0,8< -0,7;-0,6;-0,52;-0,48;-0,9;...< -0,1\)

Hay \(-\dfrac{4}{5}< -\dfrac{7}{10};-\dfrac{6}{10};-\dfrac{13}{25};-\dfrac{12}{25};-\dfrac{9}{10};...< -\dfrac{1}{10}\)

\(\Rightarrow x\in\left\{-\dfrac{7}{10};-\dfrac{6}{10};-\dfrac{13}{25};-\dfrac{12}{25};-\dfrac{9}{10};...\right\}\)

Chúc bạn học tốt!

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=2010\)

\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)\(\Rightarrow\dfrac{-1}{x+3}=\dfrac{1}{2010}\)

\(\Rightarrow x+3=-2010\)

\(\Rightarrow x=-2013\)

Vậy x = -2013

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=\dfrac{1}{2010}\)

=> \(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)

=> \(-\dfrac{1}{x+3}=\dfrac{1}{2010}\)

=> \(-2010=x+3\)

=> \(x=-2013\)

\(-\dfrac{1}{21}+\left(-\dfrac{1}{28}\right)\\ -\dfrac{1}{21}-\dfrac{1}{28}\\ =-\dfrac{4}{84}-\dfrac{3}{84}\\=-\dfrac{7}{84}\\ =-\dfrac{1}{12} \)

P/s: Lần sau ghi đề bằng công thức toán nhé bạn!

\(\dfrac{-1}{21}+\dfrac{-1}{28}=\dfrac{-1}{21}-\dfrac{1}{28}=\dfrac{-4}{84}-\dfrac{3}{84}\)

\(=\dfrac{-7}{84}=\dfrac{-1}{12}\)

Chúc bạn học tốt!!! (cái này bạn dùng máy tính được mà)

Bài 2.6:

\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{2.1}\)

\(C=-\left(\dfrac{-1}{100}+\dfrac{1}{100.99}+......+\dfrac{1}{2.1}\right)\)

\(C=-\left(\dfrac{-1}{100}+\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+....+\dfrac{1}{2}-\dfrac{1}{1}\right)\)

\(C=-1\)

Chúc bạn học tốt!!!

2.4:

\(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}-\dfrac{2}{9}\right)-\dfrac{3}{4}+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{1}{72}-\dfrac{1}{36}\right)\)

\(=\left(\dfrac{3}{9}-\dfrac{2}{9}\right)-\dfrac{3}{4}+\left(\dfrac{9}{15}+\dfrac{1}{15}\right)+\left(\dfrac{1}{72}-\dfrac{2}{72}\right)\)

\(=\dfrac{1}{9}-\dfrac{3}{4}+\dfrac{10}{15}+\left(-\dfrac{1}{72}\right)\)

\(=\dfrac{1}{9}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{72}\)

\(=\dfrac{1}{72}\)

Bài 2.4:

\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)

Bài 2.6:

\(C=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\dfrac{99}{100}=-\dfrac{98}{100}=-\dfrac{49}{50}\)

a/ \(\left(x+1\right)\left(x-2\right)< 0\)

TH1:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\) (vô lý)

TH2:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow-1< x< 2\)

Vậy.........

b/ \(\left(x-3\right)\left(x-4\right)>0\)

TH1:\(\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\)\(\Rightarrow x>4\)

TH2:\(\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\)\(\Rightarrow x< 3\)

Vậy...............

c/ \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}-\dfrac{1}{8}\)

\(\Rightarrow\dfrac{-1}{12}< x< -\dfrac{5}{48}\)

Vậy...............

Để ( x + 1 ) ( x - 2 ) < 0

=> x + 1 và x - 2 phải khác dấu mà x + 1 > x + 2

=> x + 1 dương x + 2 âm

Tức là x + 1 > 0 => x > - 1 và x - 2 < 0 => x < 2

a) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}x+1>0;x-2< 0\\x+1< 0;x-2>0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>-1;x< 2\\x< -1;x>2\end{matrix}\right.\)

\(\Rightarrow-1< x< 2\) (tm)

Vậy \(-1< x< 2.\)

b) \(\left(x-3\right)\left(x-4\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}x-3>0;x-4>0\\x-3< 0;x-4< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>3;x>4\\x< 3;x< 4\end{matrix}\right.\)

\(\Rightarrow3< x< 4\)

Vậy \(3< x< 4.\)

c) \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\Rightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)

Vậy \(\dfrac{-1}{12}< x< \dfrac{1}{8}.\)

\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)

\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)

\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)

\(A=1-1+1=1\)

\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)

\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)

\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)

\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)

\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)

\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)

\(A=-1+1=0\)

\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)

\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)

\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)

\(B=-36-\dfrac{1}{4}\)

B = \(-\dfrac{145}{4}\)

Mk trả lời câu hỏi này mà