c) x+$\frac{1}{2}$=$2^{5}$ :$2^{3}$ d) ($x+\frac{1}{2}$)$^{2}$=$\frac{4}{25}$ 26/10/2021 Bởi Reagan c) x+$\frac{1}{2}$=$2^{5}$ :$2^{3}$ d) ($x+\frac{1}{2}$)$^{2}$=$\frac{4}{25}$
Giải thích các bước giải : c, `x + 1/2 = 2^5 : 2^3` `⇔ x + 1/2 = 2^2` `⇔ x + 1/2 = 4` `⇔ x = 4 – 1/2` `⇔ x = 4 – 0,5` `⇔ x = 3,5` Vậy `x = 3,5` d, `( x + 1/2 )^2 = 4/25` ⇔ \(\left[ \begin{array}{l}x+ \dfrac{1}{2} = \dfrac{2}{5}\\x+ \dfrac{1}{2} = \dfrac{-2}{5}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x= \dfrac{-1}{10}\\x= \dfrac{-9}{10}\end{array} \right.\) Vậy `…` Bình luận
Tham khảo ` c) x+\frac{1}{2}=2^5:2^3` `⇒x+\frac{1}{2}=2^2` `⇒x+\frac{1}{2}=4` `⇒x=4-\frac{1}{2}` `⇒x=\frac{7}{2}` `d) (x+\frac{1}{2})^2=\frac{4}{25}` `⇒(x+\frac{1}{2})^2=(\frac{2}{5})^2` `⇒`\(\left[ \begin{array}{l}x+\dfrac{1}{2}=\dfrac{2}{5}\\x+\dfrac{1}{2}=\dfrac{-2}{5}\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=\dfrac{2}{5}-\dfrac{1}{2}\\x=\dfrac{-2}{5}-\dfrac{1}{2}\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=\dfrac{-1}{10}\\x=\dfrac{-9}{10}\end{array} \right.\) Vậy `x=\frac{-1}{10}` hoặc `x=\frac{-9}{10}` Bình luận
Giải thích các bước giải :
c, `x + 1/2 = 2^5 : 2^3`
`⇔ x + 1/2 = 2^2`
`⇔ x + 1/2 = 4`
`⇔ x = 4 – 1/2`
`⇔ x = 4 – 0,5`
`⇔ x = 3,5`
Vậy `x = 3,5`
d, `( x + 1/2 )^2 = 4/25`
⇔ \(\left[ \begin{array}{l}x+ \dfrac{1}{2} = \dfrac{2}{5}\\x+ \dfrac{1}{2} = \dfrac{-2}{5}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x= \dfrac{-1}{10}\\x= \dfrac{-9}{10}\end{array} \right.\)
Vậy `…`
Tham khảo
` c) x+\frac{1}{2}=2^5:2^3`
`⇒x+\frac{1}{2}=2^2`
`⇒x+\frac{1}{2}=4`
`⇒x=4-\frac{1}{2}`
`⇒x=\frac{7}{2}`
`d) (x+\frac{1}{2})^2=\frac{4}{25}`
`⇒(x+\frac{1}{2})^2=(\frac{2}{5})^2`
`⇒`\(\left[ \begin{array}{l}x+\dfrac{1}{2}=\dfrac{2}{5}\\x+\dfrac{1}{2}=\dfrac{-2}{5}\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=\dfrac{2}{5}-\dfrac{1}{2}\\x=\dfrac{-2}{5}-\dfrac{1}{2}\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=\dfrac{-1}{10}\\x=\dfrac{-9}{10}\end{array} \right.\)
Vậy `x=\frac{-1}{10}` hoặc `x=\frac{-9}{10}`