BT
NT
27 tháng 10 2023 lúc 22:13

\(\dfrac{3}{4}+\dfrac{9}{4}\left(x-\dfrac{1}{3}\right)^2=\dfrac{13}{16}\)

=>\(\dfrac{9}{4}\left(x-\dfrac{1}{3}\right)^2=\dfrac{13}{16}-\dfrac{3}{4}=\dfrac{1}{16}\)

=>\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{16}:\dfrac{9}{4}=\dfrac{1}{16}\cdot\dfrac{4}{9}=\dfrac{1}{36}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{6}\\x-\dfrac{1}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\x=\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}\end{matrix}\right.\)

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H24
27 tháng 10 2023 lúc 22:13

`3/4 + 9/4 (x-1/3)^2 =13/16`

`=>  9/4 (x-1/3)^2 =13/16-3/4`

`=>  9/4 (x-1/3)^2 =1/16`

`=> (x-1/3)^2 =1/16:9/4`

`=> (x-1/3)^2 =1/16 xx 4/9`

`=> (x-1/3)^2 =1/36`

`=> (x-1/3)^2 = (+- 1/6)^2`

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{6}\\x-\dfrac{1}{3}=-\dfrac{1}{6}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

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HD
27 tháng 10 2023 lúc 22:12

tìm x hả bn ơi

 

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