Toán a,(x – 1/4 )^2 – 25/64 = 0 b,(x- 1/4)^2 + 17/64 = 21/32 06/12/2021 By Alice a,(x – 1/4 )^2 – 25/64 = 0 b,(x- 1/4)^2 + 17/64 = 21/32
Đáp án: a.$x\in\{-\dfrac38,\dfrac78\}$ b.$x\in\{-\dfrac38,\dfrac78\}$ Giải thích các bước giải: a.$(x-\dfrac14)^2-\dfrac{25}{64}=0$ $\to (x-\dfrac14)^2=\dfrac{25}{64}$ $\to x-\dfrac14=\dfrac58\to x=\dfrac78$ Hoặc $x-\dfrac14=-\dfrac58\to x=-\dfrac38$ b.$(x-\dfrac14)^2+\dfrac{17}{64}=\dfrac{21}{32}$ $\to (x-\dfrac14)^2=\dfrac{25}{64}$ $\to x-\dfrac14=\dfrac58\to x=\dfrac78$ Hoặc $x-\dfrac14=-\dfrac58\to x=-\dfrac38$ Trả lời
$a,$ $(x – \dfrac{1}{4})^2 – \dfrac{25}{64}=0$ $⇔$ $(x – \dfrac{1}{4})^2 = \dfrac{25}{64}$ $⇔$ $x – \dfrac{1}{4}$ = (±$\dfrac{5}{8}$) $⇒$ \(\left[ \begin{array}{l}x=\dfrac{5}{8}+\dfrac{1}{4}\\x=\dfrac{-5}{8}+\dfrac{1}{4}\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=\dfrac{7}{8}\\x=\dfrac{-3}{8}\end{array} \right.\) Vậy $x$ ∈ {$\dfrac{7}{8};\dfrac{-3}{8}$} $b,$ $(x – \dfrac{1}{4})^2 + \dfrac{17}{64}=\dfrac{21}{32}$ $⇔$ $(x – \dfrac{1}{4})^2 = \dfrac{21}{32} – \dfrac{17}{64}$ $⇔$ $(x – \dfrac{1}{4})^2 = \dfrac{25}{64}$ $⇔$ $x – \dfrac{1}{4}$ = (±$\dfrac{5}{8}$) $⇒$ \(\left[ \begin{array}{l}x=\dfrac{5}{8}+\dfrac{1}{4}\\x=\dfrac{-5}{8}+\dfrac{1}{4}\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=\dfrac{7}{8}\\x=\dfrac{-3}{8}\end{array} \right.\) Vậy $x$ ∈ {$\dfrac{7}{8};\dfrac{-3}{8}$} Trả lời
Đáp án: a.$x\in\{-\dfrac38,\dfrac78\}$
b.$x\in\{-\dfrac38,\dfrac78\}$
Giải thích các bước giải:
a.$(x-\dfrac14)^2-\dfrac{25}{64}=0$
$\to (x-\dfrac14)^2=\dfrac{25}{64}$
$\to x-\dfrac14=\dfrac58\to x=\dfrac78$
Hoặc $x-\dfrac14=-\dfrac58\to x=-\dfrac38$
b.$(x-\dfrac14)^2+\dfrac{17}{64}=\dfrac{21}{32}$
$\to (x-\dfrac14)^2=\dfrac{25}{64}$
$\to x-\dfrac14=\dfrac58\to x=\dfrac78$
Hoặc $x-\dfrac14=-\dfrac58\to x=-\dfrac38$
$a,$ $(x – \dfrac{1}{4})^2 – \dfrac{25}{64}=0$
$⇔$ $(x – \dfrac{1}{4})^2 = \dfrac{25}{64}$
$⇔$ $x – \dfrac{1}{4}$ = (±$\dfrac{5}{8}$)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{5}{8}+\dfrac{1}{4}\\x=\dfrac{-5}{8}+\dfrac{1}{4}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{7}{8}\\x=\dfrac{-3}{8}\end{array} \right.\)
Vậy $x$ ∈ {$\dfrac{7}{8};\dfrac{-3}{8}$}
$b,$ $(x – \dfrac{1}{4})^2 + \dfrac{17}{64}=\dfrac{21}{32}$
$⇔$ $(x – \dfrac{1}{4})^2 = \dfrac{21}{32} – \dfrac{17}{64}$
$⇔$ $(x – \dfrac{1}{4})^2 = \dfrac{25}{64}$
$⇔$ $x – \dfrac{1}{4}$ = (±$\dfrac{5}{8}$)
$⇒$ \(\left[ \begin{array}{l}x=\dfrac{5}{8}+\dfrac{1}{4}\\x=\dfrac{-5}{8}+\dfrac{1}{4}\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=\dfrac{7}{8}\\x=\dfrac{-3}{8}\end{array} \right.\)
Vậy $x$ ∈ {$\dfrac{7}{8};\dfrac{-3}{8}$}