| 2 phần 5 - x | - 5 = -1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
UT
16 tháng 4 2021
\(a.32,5-3\cdot0,87=32,5-2,61=29,89\)
\(8,5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8,5\cdot\left(\dfrac{3}{2}+\dfrac{4}{4}\right):5\\ =8,5\cdot\left(\dfrac{6}{4}+\dfrac{4}{4}\right):5\\ =8,5\cdot\dfrac{10}{4}:5\\ =\dfrac{85}{4}:5\\ =\dfrac{17}{4}\)
\(b.30,96-6,45+14,4:3=30,96-6,45+4,8\\ =29,31\)
\(\dfrac{2}{5}\cdot\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{2}{5}\cdot\left(\dfrac{8}{10}-\dfrac{5}{10}\right)\\ =\dfrac{2}{5}\cdot\dfrac{3}{10}=\dfrac{3}{25}\)
bài 2
\(a.2,5\cdot12,5\cdot8\cdot0,4=\left(2,5\cdot0,4\right)\left(12,5\cdot8\right)\\ =1\cdot100=100\)
b,\(\dfrac{12}{15}\cdot\dfrac{5}{6}\cdot\dfrac{3}{20}\cdot\dfrac{32}{5}=\dfrac{12\cdot5\cdot3\cdot32}{15\cdot6\cdot20\cdot5}\\ =\dfrac{3\cdot4\cdot5\cdot3\cdot4\cdot8}{3\cdot5\cdot2\cdot3\cdot5\cdot4\cdot5}=\dfrac{16}{25}\)
16 tháng 4 2021
Bài 1:
a) \(32.5-3\cdot0.87=32.5-2.61=29.89\)
\(8.5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8.5\cdot\dfrac{5}{2}:5=\dfrac{17}{2}\cdot\dfrac{5}{2}:5=\dfrac{85}{4}\cdot\dfrac{1}{5}=\dfrac{17}{4}\)
TN
9 tháng 3 2020
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(|\frac{2}{5}-x|-5= -1\)
\(|\frac{2}{5}-x| = -1 + 5 \)
\(|\frac{2}{5}-x| = 4\)
Ta có 2 trường hợp :
\(\frac{2}{5}-x = -4\)
\(x=\frac{2}{5}-(-4) \)
\(x=\frac{22}{5}\)
Trường hợp 2 :
\(\frac{2}{5} -x= 4\)
\(x =\frac{2}{5} - 4\)
\(x=-\frac{18}{5}\)
Vậy : \(x=\frac{22}{5}\) hoặc \(-\frac{18}{5}\)