Bài 1: tính:
a) \(\dfrac{-2}{5}\cdot\dfrac{1}{3}-0,2\)
\(=\dfrac{-2}{15}-\dfrac{3}{15}\)
\(=\dfrac{-1}{3}\)
b) \(\dfrac{2}{3}:0,25+\dfrac{1}{5}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{10}{60}+\dfrac{12}{60}\)
\(=\dfrac{11}{30}\)
c) \(\dfrac{1}{6}\cdot\dfrac{2}{5}-\dfrac{2}{5}\cdot0,75+0,4\cdot\dfrac{1}{3}\)
\(=\dfrac{1}{6}\cdot\dfrac{2}{5}-\dfrac{2}{5}\cdot\dfrac{3}{4}+\dfrac{2}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{30}-\dfrac{9}{30}+\dfrac{4}{30}\)
\(=\dfrac{-1}{10}\)
d) \(\dfrac{2}{3}-\dfrac{1}{6}:\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{4}{6}-\dfrac{6}{6}-\dfrac{3}{6}\)
\(=\dfrac{-5}{6}\)
Làm bài 5 nha :
a) \(\dfrac{3x-1}{201}+\dfrac{3x-1}{301}=\dfrac{3x-1}{401}+\dfrac{3x-1}{501}\)
\(\Leftrightarrow\dfrac{3x-1}{201}+\dfrac{3x-1}{301}-\dfrac{3x-1}{401}-\dfrac{3x-1}{501}=0\)
\(\Leftrightarrow\left(3x-1\right)\left(\dfrac{1}{201}+\dfrac{1}{301}+\dfrac{1}{401}+\dfrac{1}{501}\right)=0\)
\(\Leftrightarrow3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy .............
b ) \(\dfrac{x-1}{99}+\dfrac{x-2}{98}=\dfrac{x-3}{97}+\dfrac{x-4}{96}\)
\(\Leftrightarrow\dfrac{x-1}{99}-\dfrac{99}{99}+\dfrac{x-2}{98}-\dfrac{98}{98}=\dfrac{x-3}{97}-\dfrac{97}{97}+\dfrac{x-4}{96}-\dfrac{96}{96}\)
\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{98}=\dfrac{x-100}{97}+\dfrac{x-100}{96}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\right)=0\)
\(\Leftrightarrow x=100.\)
c ) \(\dfrac{x+5}{2018}+\dfrac{x+6}{2017}=\dfrac{x+7}{2016}+\dfrac{x+8}{2015}\)
\(\Leftrightarrow\dfrac{x+2023}{2018}+\dfrac{x+2023}{2017}=\dfrac{x+2023}{2016}+\dfrac{x+2023}{2015}\)
\(\Leftrightarrow\left(x+2023\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow x=-2023\)
Vậy ..........
Bài 2: tìm x:
a) \(\dfrac{2}{5}-\dfrac{1}{3}:x=0,75\)
\(\Rightarrow\dfrac{1}{3}:x=\dfrac{-3}{10}\)
\(\Rightarrow x=\dfrac{-10}{9}\)
b) \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right):\dfrac{1}{3}x=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-1}{6}:\dfrac{1}{3}x=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{3}x=\dfrac{-2}{3}\)
\(\Rightarrow x=-2\)
c) \(\left(2-\dfrac{1}{3}x\right):\dfrac{1}{2}=0,25\)
\(\Rightarrow2-\dfrac{1}{3}x=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{3}x=\dfrac{15}{8}\)
\(\Rightarrow x=\dfrac{45}{8}\)
d) \(\left(3-x\right)\left(2x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow3-x=0\) hoặc \(2x+\dfrac{1}{3}=0\)
Xét \(3-x=0\Rightarrow x=3\)
Xét \(2x+\dfrac{1}{3}=0\Rightarrow2x=\dfrac{-1}{3}\Rightarrow x=\dfrac{-1}{6}\)
- Xin làm nốt bài còn lại :v
* \(x+y=xy=\dfrac{x}{y}\)
Ta có : \(x+y=xy\Leftrightarrow x=xy-y\Leftrightarrow x=y\left(x-1\right)\)
Vậy \(\dfrac{x}{y}=\dfrac{y\left(x-1\right)}{y}=x-1\) ( 1 )
Đề cho \(x+y=\dfrac{x}{y}\) theo (1) ta có :
\(x-1=x+y\Leftrightarrow x-x-y=1\Leftrightarrow y=-1\)
Thay vào ta có \(x=\dfrac{1}{2}\).
* ) \(\left(xy.yz.zx\right)^2=\dfrac{1}{25}\)
\(\Leftrightarrow xy.yz.zx=\pm\dfrac{1}{5}\)
Xảy ra 2 trường hợp :
TH1 : \(x.y.z=\dfrac{1}{5}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{5}\\y=-\dfrac{1}{2}\\z=-\dfrac{2}{3}\end{matrix}\right.\)
TH2: \(x.y.z=-\dfrac{1}{5}\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{5}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....
- Mk nhận đc rồi nhé, yêu Lan Anh nhiều
- Nhân tiện cảm ơn mấy bn đã giúp mk làm bài.
