Câu hỏi của Alana Nguyễn

Question

$2x^2+0.82=1$

$4x^2-1=0$

$\left(3x-\dfrac{1}{4}\right).\left(x+\dfrac{1}{2}\right)=0$

Tìm số chưa biết

Võ Đông Anh Tuấn · Accepted Answer

- V~V cả toán lớp 10Câu trả lời: - V~V cả toán lớp 10

Trần Nguyễn Bảo Quyên · Answer

$a.$

$2x^2+0.82=1$

$\Rightarrow2x^2+0=1$

$\Rightarrow2x^2=1$

$\Rightarrow x^2=\dfrac{1}{2}$

$\Rightarrow x=\pm\sqrt{\dfrac{1}{2}}$

$b.$

$4x^2-1=0$

$4x^2=1$

$\Rightarrow x^2=\dfrac{1}{4}$

$\Rightarrow x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}$

$c.$

$\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0$

$\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\x+\dfrac{1}{2}=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\x=-\dfrac{1}{2}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}\x=-\dfrac{1}{2}\end{matrix}\right.$Câu trả lời: $a.$

$2x^2+0.82=1$

$\Rightarrow2x^2+0=1$

$\Rightarrow2x^2=1$

$\Rightarrow x^2=\dfrac{1}{2}$

$\Rightarrow x=\pm\sqrt{\dfrac{1}{2}}$

$b.$

$4x^2-1=0$

$4x^2=1$

$\Rightarrow x^2=\dfrac{1}{4}$

$\Rightarrow x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}$

$c.$

$\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0$

$\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\x+\dfrac{1}{2}=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\x=-\dfrac{1}{2}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{12}\x=-\dfrac{1}{2}\end{matrix}\right.$

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