||2x-1|-1/2|=1/5 giúp mk với

Question

||2x-1|-1/2|=1/5 giúp mk với

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1 year 2020-10-20T23:20:20+00:00 2 Answers 122 views 0

1. $\huge\text{|}$|2x-1|-1/2$\huge\text{|}$=1/5

⇔ $$\left[ \begin{array}{l}|2x-1|-\dfrac{1}{2}=\dfrac{1}5\\|2x-1|-\dfrac{1}{2}=-\dfrac{1}5\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}|2x-1|=\dfrac{1}5+\dfrac{1}{2}\\|2x-1|=-\dfrac{1}5+\dfrac{1}{2}\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}|2x-1|=\dfrac{7}{10}\\|2x-1|=\dfrac{3}{10}\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}2x-1=\dfrac{7}{10} \\2x-1=-\dfrac{7}{10}\end{array} \right.$$ hoặc $$\left[ \begin{array}{l}2x-1=\dfrac{3}{10}\\2x-1=-\dfrac{3}{10}\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}2x=\dfrac{7}{10}+1 \\2x=-\dfrac{7}{10}+1\end{array} \right.$$ hoặc $$\left[ \begin{array}{l}2x=\dfrac{3}{10}+1\\2x=-\dfrac{3}{10}+1\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}2x=\dfrac{17}{10} \\2x=-\dfrac{17}{10}\end{array} \right.$$ hoặc $$\left[ \begin{array}{l}2x=\dfrac{13}{10}\\2x=-\dfrac{13}{10}\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}x=\dfrac{17}{20} \\2x=-\dfrac{17}{20}\end{array} \right.$$ hoặc $$\left[ \begin{array}{l}x=\dfrac{13}{20}\\2x=-\dfrac{13}{20}\end{array} \right.$$

Vậy x=17/20 hoặc x=- 17/20 hoặc x=13/20 hoặc x=- 13/20

Học tốt !

2. TH: là trường hợp lớn, Th: là trường hợp nhỏ.

$||2x-1|-\dfrac{1}{2}|=\dfrac{1}{5}$

TH1: $|2x-1|-\dfrac{1}{2}=\dfrac{1}{5}$

$|2x-1|=\dfrac{1}{5}+\dfrac{1}{2}=\dfrac{7}{10}$

Th1: $2x-1=\dfrac{7}{10}$

$2x=\dfrac{7}{10}+1=\dfrac{17}{10}$

$x=\dfrac{17}{10}:2=\dfrac{17}{20}$

Th2: $2x-1=-\dfrac{7}{10}$

$2x=-\dfrac{7}{10}+1=\dfrac{3}{10}$

$x=\dfrac{3}{10}:2=\dfrac{3}{20}$

TH2: $|2x-1|-\dfrac{1}{2}=-\dfrac{1}{5}$

$|2x-1|=-\dfrac{1}{5}+\dfrac{1}{2}=\dfrac{3}{10}$

Th1: $2x-1=\dfrac{3}{10}$

$2x=\dfrac{3}{10}+1=\dfrac{13}{10}$

$x=\dfrac{13}{10}:2=\dfrac{13}{20}$

Th2: $2x-1=-\dfrac{3}{10}$

$2x=-\dfrac{3}{10}+1=\dfrac{7}{10}$

$x=\dfrac{7}{10}:2=\dfrac{7}{20}$