Tìm X:
X * 6 + x : 3 = 1/2
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\(\Leftrightarrow x^3+2x^2-3x-x^3-3x^2=-4\)
\(\Leftrightarrow x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>x=-4 hoặc x=1
1) \(x:\dfrac{1}{3}=\dfrac{1}{2}+\dfrac{1}{3}\)
\(\Rightarrow3\times x=\dfrac{5}{6}\Rightarrow x=\dfrac{5}{18}\)
2) \(\left(x:\dfrac{2}{3}\right):\dfrac{2}{5}=\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{2}\times x=\dfrac{7}{10}\times\dfrac{2}{5}=\dfrac{7}{25}\)
\(\Rightarrow x=\dfrac{7}{25}:\dfrac{3}{2}=\dfrac{14}{75}\)
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: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
\(x\times\dfrac{4}{5}=3-\dfrac{2}{5}\)
\(x\times\dfrac{4}{5}=\dfrac{13}{5}\)
\(x=\dfrac{13}{5}:\dfrac{4}{5}\)
\(x=\dfrac{13}{5}\times\dfrac{5}{4}\)
\(x=\dfrac{13}{4}\)
Chiều dài HCN là :
\(\dfrac{5}{6}:\dfrac{2}{3}=\dfrac{5}{4}\left(m\right)\)
Chu vi HCN là :
\(\left(\dfrac{5}{4}+\dfrac{2}{3}\right)\times2=\dfrac{23}{6}\left(m\right)\)
Đ/s :...
\(x\times\dfrac{4}{5}=\dfrac{13}{5}\\ x=\dfrac{13}{5}\div\dfrac{4}{5}\\ x=\dfrac{13}{4}\)
\(CD=\dfrac{5}{6}\div\dfrac{2}{3}=\dfrac{5}{4}\\ P=\left(\dfrac{5}{4}+\dfrac{2}{3}\right)\times2=\dfrac{23}{6}\)
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
Ta có PT: 6x+x/3=1/2
Giải PT ta đc x=3/38
Nhớ k giùm mk nha!