Tìm x:
x = 1 + 1 + 3
10 2 5 10
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)
\(x=\dfrac{-42}{20}+\dfrac{5}{20}=\dfrac{-37}{20}\)
1) \(x:\dfrac{1}{3}=\dfrac{1}{2}+\dfrac{1}{3}\)
\(\Rightarrow3\times x=\dfrac{5}{6}\Rightarrow x=\dfrac{5}{18}\)
2) \(\left(x:\dfrac{2}{3}\right):\dfrac{2}{5}=\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{2}\times x=\dfrac{7}{10}\times\dfrac{2}{5}=\dfrac{7}{25}\)
\(\Rightarrow x=\dfrac{7}{25}:\dfrac{3}{2}=\dfrac{14}{75}\)
\(\Leftrightarrow x+\dfrac{1}{8}=\dfrac{1}{6}\\ \Leftrightarrow x=\dfrac{1}{24}\)
\(\Rightarrow x+\dfrac{1}{8}=\dfrac{1}{6}\\ \Rightarrow x=\dfrac{1}{24}\)
Bài 1:
\(=\dfrac{-3-39}{42}+\dfrac{-6-11}{17}-\dfrac{1}{6}=-\dfrac{119}{48}\)
Bài 2:
=>x:5=-13/20
hay x=-65/20=-13/4
1.
\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)
2.
\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt
Bài 1:
Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=6x\left(2x+1\right)\cdot2y\)
\(=12xy\left(2x+1\right)\)
Bài 2:
Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)
=>(x+1)2=16
=>x+1=\(\sqrt{16}\)
=>x+1=4
=>x=3
À, nhầm :v
\(x=\frac{1}{10}+\frac{1}{2}+\frac{1}{5}+\frac{3}{10}\)
\(x=\left(\frac{3}{10}+\frac{1}{10}\right)+\frac{1}{2}+\frac{1}{5}\)
\(x=\frac{11}{10}\)
x = 1 + 1 + 3
x = 5
=> x = 5