\(\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\) (nhận)
hoặc \(\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\) (nhận)
Vậy x = 0; x = 4
Tìm x:
1) \(\text{(x−1):0,16=−9:(1−x)}\)
2) \(\left(\left|x\right|-\dfrac{3}{2}\right)\left(2x^2-10\right)=0\)
3)\(8\sqrt{x}=x^2\left(x\ge0\right)\)
\(x-2\times\sqrt{x}=0\left(x\ge0\right)\)
tìm số hữu tỉ x biết: \(x-2\sqrt{x}=0\left(x\ge0\right)\))
Tìm số hữu tỉ x, biết:
\(x-2\sqrt{x}=0\left(x\ge0\right)\)
a) Tìm x biết:\(x-2\sqrt{x}=0\left(x\ge0\right)\)
tìm x biết
a)\(\frac{3.\left(x-1\right)}{2}=\frac{8}{27\left(x-1\right)}\)
b)\(x-3\sqrt{x}=0\) với \(x\ge0\)
\(\left(2x+3\right)^2+\left(3x-2\right)^4=0\) vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\) nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\) \(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)
Tìm x
\(8\sqrt{x}=x^2\left(x\ge0\right)\)
tim x
a, \(\left(x+1\right)\left(2-x\right)\left(3+x^2\right)\ge0\)
b, \(\left(2x+1\right)\left(x+3\right)\left(x-5\right)>0\)
c,\(\frac{\left(3x-6\right)\left(x-4\right)}{x+3}< 0\)
