Câu hỏi của khánh chi nguyễn

Question

a,1/2(x-4)-1/4(x-4/3)=2(x-1/2)

b, 3/4-(x-1/2)^2=-11/2

c,3/16+1/1/16.(x-2/3)^2=3/4

『Kuroba ム Tsuki Ryoo... · Accepted Answer

`#040911`a,$\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)$$\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\
\Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\
\Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\
\Rightarrow x=-\dfrac{8}{21}$Vậy, $x=-\dfrac{8}{21}$b,$\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}$$\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\
\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\
\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2$$\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=3\x=-2\end{matrix}\right.$Vậy, $x\in\left\{-2;3\right\}$c,$\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}$$\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\
\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\
\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\
\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}$Bạn xem lại đề có sai kh nhỉ?Câu trả lời: `#040911`a,$\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)$$\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\
\Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\
\Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\
\Rightarrow x=-\dfrac{8}{21}$Vậy, $x=-\dfrac{8}{21}$b,$\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}$$\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\
\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\
\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2$$\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\
\Rightarrow\left[{}\begin{matrix}x=3\x=-2\end{matrix}\right.$Vậy, $x\in\left\{-2;3\right\}$c,$\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}$$\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\
\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\
\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\
\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}$Bạn xem lại đề có sai kh nhỉ?

Nguyễn Đức Trí · Answer

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

$\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}$

$\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}$

$\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16$

$\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2$

$\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\x=\dfrac{23}{48}\end{matrix}\right.$Câu trả lời: c) $\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}$