Tìm x , biết
x : 4 = 100
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\(\dfrac{x^2}{20}=\dfrac{4}{5}\)
\(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?
Ta có:
\(x^4=y^4\)
\(\Rightarrow x^4-y^4=0\)
\(\Rightarrow\left(x^2\right)^2-\left(y^2\right)^2=0\)
\(\Rightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-y^2=0\\x^2+y^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
_______________
Ta có:
\(x^5=y^5\)
\(\Rightarrow x^5-y^5=0\)
\(\Rightarrow x-y=0\)
\(\Rightarrow x=y\)
\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)
\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)
\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)
\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)
1) \(\Rightarrow x^2\left(x^{2004}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{2004}=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
2) \(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
\(x^{20}=x\\ \Leftrightarrow x^{20}-x=0\\ \Leftrightarrow x\left(x^{19}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
X:4= 100
X = 100 x 4
X = 400
x : 4 = 100
x = 100 x 4
x = 400