Question

[nguyễn lê bảo trâm]

Tìm các giá trị của biến để :

\(9.x^2-36=0\)

\(\left(x-1\right).\left(x+1\right).\left(x^2+\dfrac{1}{2}\right)=0\)

\(\left|x+4\right|+5=0\)

\(\sqrt{2.x}-3-1=0\)

Mn làm cho mik dò điii

Answer 1

[Thu Thao]

a/ \(\Leftrightarrow9x^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)

\(\Leftrightarrow x=\pm2\)

b/ \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\) (do \(x^2+\dfrac{1}{2}>0\))

\(\Leftrightarrow x=\pm1\)

c/ Có \(\left|x+4\right|\ge0\forall x\)

=> \(\left|x+4\right|+5\ge5>0\forall x\)

\(\Rightarrow\left|x+4\right|+5=0\left(vô-lí\right)\)

\(\Rightarrow x\in\varnothing\)

d/ \(\sqrt{2x}-3-1=0\)

\(\Leftrightarrow\sqrt{2x}=4\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Answer 2

[nguyễn lê bảo trâm]

thank you