Câu hỏi của Trương Thị Kiều Oanh

Question

Tìm x biết

a)$\dfrac{x}{9}=\dfrac{16}{x}=$     b)$x^3+27=0$    c)$\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x$    d)$\left|x+3\right|+\left|x-5\right|=8$

Nguyễn Thanh Hằng · Accepted Answer

a/ $\dfrac{x}{9}=\dfrac{16}{x}$

$\Leftrightarrow x^2=9.16$

$\Leftrightarrow x^2=144$

$\Leftrightarrow\left[{}\begin{matrix}x=12\x=-12\end{matrix}\right.$

Vậy ...

b/ $x^3+27=0$

$\Leftrightarrow x^3=-27$

$\Leftrightarrow x^3=\left(-3\right)^3$

$\Leftrightarrow x=-3$

Vậy ...

c/ $\left|x\left(x^2-\dfrac{5}{4}\right)=x\right|$

$\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{5}{4}\right)=x\x\left(x^2-\dfrac{5}{4}\right)=-x\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{5}{4}x=x\x^3-\dfrac{5}{4}x=-x\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^3-\left(\dfrac{5}{4}x+x\right)=0\x^3-\left(\dfrac{5}{4}x-x\right)=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{9}{4}x=0\x^3-\dfrac{1}{4}x=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{9}{4}\right)=0\x\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\x^2-\dfrac{9}{4}=0\end{matrix}\right.\\left[{}\begin{matrix}x=0\x^2-\dfrac{1}{4}=0\end{matrix}\right.\end{matrix}\right.$

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

Vậy ...

c/ Với mọi x ta có :

$\left|x-5\right|=\left|5-x\right|$

$\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge8$

Dấu "=" xảy ra khi :

$\left(x+3\right)\left(5-x\right)\ge0$

$\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\5-x\ge0\end{matrix}\right.\\left\{{}\begin{matrix}x+3\le0\5-x\le0\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\5\ge x\end{matrix}\right.\\left\{{}\begin{matrix}x\le-3\5\le x\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}-3\le x\le5\x\in\varnothing\end{matrix}\right.$

Vậy ...Câu trả lời: a/ $\dfrac{x}{9}=\dfrac{16}{x}$

$\Leftrightarrow x^2=9.16$

$\Leftrightarrow x^2=144$

$\Leftrightarrow\left[{}\begin{matrix}x=12\x=-12\end{matrix}\right.$

Vậy ...

b/ $x^3+27=0$

$\Leftrightarrow x^3=-27$

$\Leftrightarrow x^3=\left(-3\right)^3$

$\Leftrightarrow x=-3$

Vậy ...

c/ $\left|x\left(x^2-\dfrac{5}{4}\right)=x\right|$

$\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{5}{4}\right)=x\x\left(x^2-\dfrac{5}{4}\right)=-x\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{5}{4}x=x\x^3-\dfrac{5}{4}x=-x\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^3-\left(\dfrac{5}{4}x+x\right)=0\x^3-\left(\dfrac{5}{4}x-x\right)=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{9}{4}x=0\x^3-\dfrac{1}{4}x=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{9}{4}\right)=0\x\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\x^2-\dfrac{9}{4}=0\end{matrix}\right.\\left[{}\begin{matrix}x=0\x^2-\dfrac{1}{4}=0\end{matrix}\right.\end{matrix}\right.$

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

Vậy ...

c/ Với mọi x ta có :

$\left|x-5\right|=\left|5-x\right|$

$\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge8$

Dấu "=" xảy ra khi :

$\left(x+3\right)\left(5-x\right)\ge0$

$\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\5-x\ge0\end{matrix}\right.\\left\{{}\begin{matrix}x+3\le0\5-x\le0\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\5\ge x\end{matrix}\right.\\left\{{}\begin{matrix}x\le-3\5\le x\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}-3\le x\le5\x\in\varnothing\end{matrix}\right.$

Vậy ...

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