Câu hỏi của Kim So Huyn

Question

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

Hắc Hường · Accepted Answer

Giải:

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

$\Leftrightarrow\left|\left(-2\dfrac{2}{3}\right)^2-x\right|=\dfrac{1}{3}$

$\Leftrightarrow\left|\dfrac{16}{9}-x\right|=\dfrac{1}{3}$

$\Leftrightarrow\left[{}\begin{matrix}\dfrac{16}{9}-x=\dfrac{1}{3}\\dfrac{16}{9}-x=-\dfrac{1}{3}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\x=-\dfrac{19}{9}\end{matrix}\right.$Câu trả lời: Giải:

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

$\Leftrightarrow\left|\left(-2\dfrac{2}{3}\right)^2-x\right|=\dfrac{1}{3}$

$\Leftrightarrow\left|\dfrac{16}{9}-x\right|=\dfrac{1}{3}$

$\Leftrightarrow\left[{}\begin{matrix}\dfrac{16}{9}-x=\dfrac{1}{3}\\dfrac{16}{9}-x=-\dfrac{1}{3}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{9}\x=-\dfrac{19}{9}\end{matrix}\right.$

Nguyễn Tú Quyên · Answer

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

$\left(\dfrac{-10}{3}\right)^2$-x=0$\times$$-\dfrac{1}{3}$

-$\dfrac{100}{9}$-x=0

x=-$\dfrac{100}{9}$-0

x=-$\dfrac{100}{9}$Câu trả lời: $\left(-2\dfrac{2^{ }}{3}\right)^2$-x$\div$-$\dfrac{1}{3}$=0

$\left(\dfrac{-10}{3}\right)^2$-x=0$\times$$-\dfrac{1}{3}$

-$\dfrac{100}{9}$-x=0

x=-$\dfrac{100}{9}$-0

x=-$\dfrac{100}{9}$

Violympic toán 7