Bài 3: Nhân, chia số hữu tỉ

\(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)

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

=> \(-\dfrac{2}{3}< x< 3\)

p/s : Không biết đúng hay sai bởi vì đang áp dụng cách mới :v

\(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)

TH1:

\(\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow x>3\)

TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{2}{3}\)

Vậy để \(\left(x-3\right)\left(x+\dfrac{2}{3}\right)>0\)

thì \(x>3\) hoặc \(x< -\dfrac{2}{3}\)

*\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy phương trình đã cho có \(S=\left\{0;\dfrac{1}{7}\right\}\)

* \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4x}=\dfrac{2}{5}\)

\(\Leftrightarrow15x+1=8x\)

\(\Leftrightarrow7x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{7}\left(tmđk\right)\)

Vậy phương trình đã cho có \(S=\left\{\dfrac{-1}{7}\right\}\)

a, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\dfrac{1}{7}\right\}\)

b, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\Rightarrow x=\dfrac{-5}{7}\)

Vậy \(x=\dfrac{-5}{7}\)

Chúc bạn học tốt!!!

a) \(2x.\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

Chúc bạn học tốt

\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{-\dfrac{7}{40}+\dfrac{5}{11}}{-\dfrac{369}{880}}\)

\(=x+\dfrac{\dfrac{123}{440}}{\dfrac{369}{880}}\)

\(=x-\dfrac{2}{3}\)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức B.

Ta có: \(-\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{3}{3}=-1\)

Vậy giá trị biểu thức B tại \(x=-\dfrac{1}{3}\) là -1.

a,\(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}\right).\dfrac{5}{4}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right).\dfrac{5}{4}\)

\(=\left(\dfrac{-2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right).\dfrac{5}{4}\)

\(=\left(-1+1\right).\dfrac{5}{4}=0\)

Chúc bạn học tốt!!!

Câu b làm tương tự!

a) Đoàn Đức Hiếu

b) \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\left(-\dfrac{3}{22}\right)+\dfrac{5}{9}:\left(-\dfrac{3}{5}\right)\)

\(=-\dfrac{5}{9}\cdot\dfrac{22}{3}-\dfrac{5}{9}\cdot\dfrac{5}{3}\)

\(=-\dfrac{110}{27}-\dfrac{25}{27}\)

\(=-5\)

a )

\(4\dfrac{5}{9}:2\dfrac{5}{8}-7=-\dfrac{995}{189}=-5\) ( đã làm tròn )

\(\left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\) \(=\dfrac{2}{5}=0\) ( đã làm tròn )

Ta có :

\(-5< x< 0\)

\(\Rightarrow x=\left\{-4;-3;-2;-1\right\}\)

b )

\(-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)=-\dfrac{20}{21}=-1\) ( đã làm tròn )

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{7}{12}=1\) ( đã làm tròn )

Ta có : \(-1< x+0,6< 1\)

\(\Rightarrow x=\left\{-1;0\right\}\)

a) \(4\dfrac{5}{9}:2\dfrac{5}{8}-7< x< \left(3\dfrac{1}{5}:3,2+4,5\cdot1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>4\dfrac{5}{9}:2\dfrac{5}{8}-7\\x< \left(3\dfrac{1}{5}:3,2+4,5\cdot1\dfrac{31}{45}\right):\left(-21\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{227}{36}\\x< -\dfrac{2}{5}\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{2}{5}>x>-\dfrac{227}{36}\)

b) \(-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)< x+\dfrac{4}{7}< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{7}>-\dfrac{17}{21}:\left(\dfrac{5}{4}-\dfrac{2}{5}\right)\\x+\dfrac{4}{7}< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-\dfrac{32}{21}\\x< \dfrac{1}{84}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{84}>x>-\dfrac{32}{21}\)

a, \(A=x^2+4x\)

Để A đạt giá trị dương thì \(x^2+4x>0\Rightarrow x.\left(x+4\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\)

+, Xét \(\left\{{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>-4\end{matrix}\right.\)

\(\Rightarrow x>0\)

+, Xét \(\left\{{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 0\\x< -4\end{matrix}\right.\)

\(\Rightarrow x< -4\)

Vậy...............

Các câu còn lại làm tương tự! Chúc bạn học tốt!!!

\(a,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\)

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

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

\(b,\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)\)

\(=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)\)

\(=\dfrac{11}{20}.\dfrac{-2}{5}\)

\(=-\dfrac{11}{50}\)

a) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Leftrightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=-\dfrac{1}{4}.\dfrac{20}{7}=\dfrac{-5}{7}\)

b)\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(\Leftrightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{5}{-7}\)
\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow2x=0\Rightarrow x=0\)

\(\Leftrightarrow x-\dfrac{1}{7}=0\Rightarrow x=\dfrac{1}{7}\)

b, 2x.(x-\(\dfrac{1}{7}\) )=0

x(2-\(\dfrac{1}{7}\))=0

x.\(\dfrac{13}{7}\)=0

x=0:\(\dfrac{13}{7}\)

x=0

Vậy x=0

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{9.10}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(A=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(B=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(B=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-.....+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\)\(B=0-1=-1\)

\(B=\left(3\dfrac{1}{5}:3,2+4,5.1\dfrac{32}{45}\right):-21\dfrac{1}{2}\)

\(B=\left(3,2:3,2+7,7\right):-21,5\)

\(B=\left(1+7,7\right):-21,5=8,7:-21,5=\dfrac{-87}{215}\)

Chúc bạn học tốt!!!