5-(1,3-x)^2=1
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Akai Haruma
Giáo viên
18 tháng 12 2023
Lời giải:
$x: \frac{5}{6}=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$
$x=\frac{1}{4}\times \frac{5}{6}=\frac{5}{24}$
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$x:1,3=83,4+6,78=90,18$
$x=90,18\times 1,3=117,234$
18 tháng 12 2023
x:5/6+1/2=3/4 x:1,3-6,78=83,4
X:5/6 =3/4-1/2 x:1,3 =83,4+6,78
x:5/6 =1/4 x:1,3 =90,18
X =1/4*5/6 x =90,18x1,3
x =117,234
X =5/24
NA
0
1 tháng 10 2021
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
24 tháng 12 2023
a: \(12:\dfrac{3}{4}+75\%\cdot\dfrac{1}{2}-16\cdot50\%\)
\(=12\cdot\dfrac{4}{3}+\dfrac{3}{4}\cdot\dfrac{1}{2}-16\cdot0,5\)
\(=16-8+\dfrac{3}{8}\)
\(=8,375\)
b: \(75\%\cdot4+22,5:3-1\dfrac{3}{5}\)
\(=0,75\cdot4+7,5-1,6\)
\(=3+7,5-1,6\)
=10,5-1,6
=8,9
5-(1,3-x)^2=1
=>(1,3-x)^2=4
=>\(\orbr{\begin{cases}1,3-x=2&1,3-x=-2&\end{cases}=>\orbr{\begin{cases}x=-0,7\\x=-3,3\end{cases}}}\)
Ta có: \(5-\left(1,3-x\right)^2=1\)
\(\Rightarrow\left(1,3-x\right)^2=2^2\)
\(\Rightarrow1,3-x=2\)
\(\Rightarrow x=1,3-2\)
\(\Rightarrow x=\frac{-10}{7}\)