Cậu ấy gửi đề mà các cậu @@

nag nỏ gửi trên face ak

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 ....

Nguyễn Đình Tân chời ơi ko để ý ở trên :v

Kệ làm hộ cho mấy cậu ấy luôn, tích chút công đức

- Mk nhận đc rồi nhé, yêu Lan Anh nhiều

- Nhân tiện cảm ơn mấy bn đã giúp mk làm bài.

ham a dau co bao ông làm đâu.............@@@@@

Wendy Thảo Quên nhìn tưởng đăng hỏi bài -_-

gửi cho ai thế bn Wendy Thảo

bn cùng trường

học thêm 1 lớp

nick câu ấy là ... trần khánh chi