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AN
27 tháng 10 2017
Tìm X
a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)
\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)
\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)
\(\Leftrightarrow\dfrac{5}{32}-x=0\)
\(\Leftrightarrow x=\dfrac{5}{32}\)
DK
0
23 tháng 12 2023
\(\dfrac{3}{5}x-\dfrac{25}{4}=\dfrac{1}{2}:x+\dfrac{5}{4}\)
=>\(\dfrac{3}{5}x-\dfrac{25}{4}-\dfrac{1}{2x}+\dfrac{5}{4}\)
=>\(\dfrac{3}{5}x-\dfrac{1}{2x}=\dfrac{30}{4}\)
=>\(\dfrac{3x}{5}-\dfrac{1}{2x}=\dfrac{30}{4}\)
=>\(\dfrac{6x^2-5}{10x}=\dfrac{30}{4}\)
=>\(6x^2-5=10x\cdot\dfrac{30}{4}=5x\cdot15=75x\)
=>\(6x^2-75x-5=0\)
=>\(x=\dfrac{75\pm\sqrt{5745}}{12}\)
22 tháng 7 2017
a)
\(3:\left(\dfrac{9}{4}\right)=\dfrac{3}{4}:\left(6.x\right)\\ \Rightarrow3.6.x=\dfrac{3}{4}.\dfrac{9}{4}\\ x=\dfrac{3}{4}.\dfrac{9}{4}.\dfrac{1}{3}.\dfrac{1}{6}\\ x=\dfrac{3}{4.4.2}\\ x=\dfrac{3}{32}\)
b)
\(4,5:0,3=\left(5.0,09\right):\left(0,01.x\right)\\ 0,01.x.4,5=5.0,09.0,3\\ x=5.\dfrac{9}{100}.\dfrac{3}{10}.100.\dfrac{10}{45}\\ x=3\)
d)
\(\left(\dfrac{1}{9}.x\right)=\dfrac{7}{4}:\dfrac{2}{25}\\ \left(\dfrac{1}{9}.x\right)=\dfrac{7}{4}.\dfrac{25}{2}\\ x:\dfrac{7}{4}=\dfrac{25}{2}:\dfrac{1}{9}\\ x=\dfrac{25}{2}.9.\dfrac{7}{4}\\ x=\dfrac{1575}{8}\\ x=196\dfrac{7}{8}\)
e)
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\\ -x.x=-2.\dfrac{8}{25}\\ -x^2=-\dfrac{16}{25}=-\dfrac{4^2}{5^2}\\ -x^2=-\left(\dfrac{4}{5}\right)^2\\ \Rightarrow x=\dfrac{4}{5}\)
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NH
1
3 tháng 12 2023
ĐKXĐ: \(x>=-\dfrac{1.96}{3}=-\dfrac{196}{300}=\dfrac{-49}{75}\)
\(\sqrt{\dfrac{1.96+3x}{4}}=\sqrt{0,04}+\dfrac{1}{4}\cdot\sqrt{\dfrac{256}{25}}\)
=>\(\sqrt{\dfrac{3x+1.96}{4}}=0.2+\dfrac{1}{4}\cdot\dfrac{16}{5}=1\)
=>\(\dfrac{3x+1,96}{4}=1\)
=>3x+1,96=4
=>3x=2,04
=>\(x=\dfrac{2.04}{3}=0.68\left(nhận\right)\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 7 2023
(\(x\) + 1)2 = \(\dfrac{4}{25}\)
(\(x+1\))2 = (\(\dfrac{2}{5}\))2
\(\left[{}\begin{matrix}x+1=-\dfrac{2}{5}\\x+1=\dfrac{2}{5}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)
Vậy \(x\in\){ \(-\dfrac{7}{5}\) ; - \(\dfrac{3}{5}\)}
4 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`(x+1)^2 = 4/25`
`=> (x+1)^2 = (+-2/5)^2`
`=>`\(\left[{}\begin{matrix}x+1=\dfrac{2}{5}\\x+1=-\dfrac{2}{5}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{2}{5}-1\\x=-\dfrac{2}{5}-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{7}{5}\end{matrix}\right.\)
Vậy, `x \in {-3/5; -7/5}.`
ND
Nguyễn Đức Trí
VIP
12 tháng 7 2023
a) \(x=\dfrac{1}{3}-\dfrac{5}{6}+\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{10}{12}+\dfrac{3}{12}=-\dfrac{3}{12}=-\dfrac{1}{4}\)
b) \(x-\dfrac{3}{25}=\dfrac{1}{10}-\dfrac{1}{50}=\dfrac{5}{50}-\dfrac{1}{50}=\dfrac{4}{50}=\dfrac{2}{25}\)
\(x=\dfrac{2}{25}+\dfrac{3}{25}=\dfrac{5}{25}=\dfrac{1}{5}\)
13 tháng 10 2021
a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)
\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)
26 tháng 10 2015
\(\left(x-\frac{1}{4}\right)^2=\left(\frac{1}{5}\right)^2=>x-\frac{1}{4}=\frac{1}{5}=>x=\frac{1}{5}+\frac{1}{4}=>x=\frac{9}{20}\)
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