tìm x biết x-2=3/8
giúp mik với
K
Khách
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Những câu hỏi liên quan
13 tháng 11 2021
\(x\left(x-2\right)+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)
\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)
5 tháng 11 2021
\(=x^2+6x+9-17=\left(x+3\right)^2-17=\left(x+3-\sqrt{17}\right)\left(x+3+\sqrt{17}\right)\)
VN
0
LA
1
15 tháng 7 2021
Ta có: \(3x^2-x+2\)
\(=3\left(x^2-\dfrac{1}{3}x+\dfrac{2}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)
\(=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}>0\forall x\)(đpcm)
NH
3
NA
20 tháng 1 2022
\(\Rightarrow a^2-1⋮8\\ \Rightarrow a^2⋮9\\ \Rightarrow a=\pm3\)
NC
1
LD
2 tháng 1 2022
27:(x-3/2)^3=(x-3/2):3
Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81
\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)
N
1
\(x-2=\frac{3}{8}\)
\(x=\frac{3}{8}+2\)
\(x=\frac{3+2\times8}{8}\)
\(x=\frac{19}{8}\)
\(x-2=\frac{3}{8}\)
\(x=\frac{3}{8}+2\)
\(x=\frac{19}{8}\)