Tìm x biết
\(x\left(x+1\right)=0\)
\(3x\left(2x-1\right)=0\)
\(\left(x+1\right)\cdot\left(x-2\right)=0\)
\(x^2\cdot\left(x+4\right)=0\)
\(\left(x+1\right)^2\cdot\left(3x-5\right)=0\)
\(x^2+1=0\)
\(3x^2\cdot\left(2x-5\right)^2=0\)
\(1\cdot2\cdot3\cdot.....\cdot100x=0\)
\(\left(\frac{3}{4}\right)^x=1\)
\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}\)
a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)
h)Tương tự các câu trên
i) x = 0
k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)
l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)
=> x + 1 = 3 => x = 2
x.(x+1)=0
suy ra x=0 hoac x+1=0
x=0-1
x=-1
vay x=0 hoac x=-1
mấy câu sau cũng làm tương tự
