Tìm x \(\in\) , biết |\(\dfrac{1}{4}\) + x |= \(\dfrac{5}{6}\)
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25 tháng 8 2023
a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
23 tháng 6 2022
\(\Leftrightarrow\left(19.75\right):x=\left(\dfrac{33}{5}-\dfrac{51}{16}\right)\cdot\dfrac{35}{6}:\dfrac{5}{2}\)
\(\Leftrightarrow19.75:x=\dfrac{637}{80}\)
hay x=1580/637
VM
7 tháng 3 2017
Quy đồng, ta được:
\(\dfrac{0,8:\left(\dfrac{4}{5}.\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{500-2}{5}\right):\dfrac{4}{7}}{6\left(\dfrac{5}{9}-\dfrac{13}{4}\right).\dfrac{36}{17}}\)
\(=\dfrac{0,8:1}{\dfrac{15}{25}}+\dfrac{\dfrac{498}{5}.\dfrac{7}{4}}{6\left(\dfrac{20-117}{36}\right).\dfrac{36}{17}}\)
\(=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}+\dfrac{\dfrac{1743}{10}}{6.\dfrac{-97}{36}.\dfrac{36}{17}}\)
\(=\dfrac{4}{3}+\dfrac{\dfrac{1743}{10}}{\dfrac{-582}{17}}\)
\(=\dfrac{4}{3}-\dfrac{9877}{1940}=\dfrac{4.1940-9877.3}{3.1940}=\dfrac{-21871}{5820}\)
Số to quá !!!!
NM
2
24 tháng 8 2023
\(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right):\left(x-\dfrac{13}{7}\right)< 0\)
\(TH1:\left\{{}\begin{matrix}x-\dfrac{1}{5}>0\\x-\dfrac{13}{7}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{13}{7}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{5}< x< \dfrac{13}{7}\)
\(TH2:\left\{{}\begin{matrix}x-\dfrac{1}{5}< 0\\x-\dfrac{13}{7}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\) (vô lý nên loại)
Vậy \(\dfrac{1}{5}< x< \dfrac{13}{7}\) thỏa mãn đề bài
NP
1
9 tháng 7 2017
Tìm x biết :
\(\dfrac{x+5}{-12}\) = \(\dfrac{-12}{x+5}\)
=> (x+5) . (x+5) = (-12) . (-12)
=> x + 5 = 12 hoặc x + 5 = -12
*Nếu x +5 = 12 => x = 7
*Nếu x + = -12 => x = -17
Vậy x = 7 hoặc x= -17
HV
11
17 tháng 8 2018
a) điều kiện xác định : \(x\ge0;x\ne1\)
ta có : \(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
b) để \(P=\dfrac{2}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left(\sqrt{x}-1\right)=2\left(x+\sqrt{x}+1\right)\Leftrightarrow3\sqrt{x}-3=2x+2\sqrt{x}+2\)
\(\Leftrightarrow2x-\sqrt{x}+5=0\Leftrightarrow2\left(x-\dfrac{1}{2}\sqrt{x}+\dfrac{1}{16}\right)+\dfrac{79}{16}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{4}\right)^2+\dfrac{79}{16}=0\left(vôlí\right)\)
vậy không tồn tại \(x\) để \(P=\dfrac{2}{3}\)
LD
1
20 tháng 8 2023
\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)
\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)
\(\Rightarrow2xy=54+y\)
\(\Rightarrow2xy-y=54\)
\(\Rightarrow xy-\dfrac{y}{2}=27\)
\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)
\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)
27 tháng 6 2022
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}-\dfrac{1213}{100}=2\cdot\left[\left(x-\dfrac{10}{7}\right)\cdot\dfrac{49}{50}+\dfrac{2}{5}\right]\)
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}=\dfrac{49}{25}\left(x-\dfrac{10}{7}\right)+\dfrac{4}{5}+\dfrac{1213}{100}\)
\(\Leftrightarrow\dfrac{1313}{100}-\dfrac{101}{25}x=\dfrac{49}{25}x-\dfrac{490}{175}+\dfrac{1293}{100}\)
=>-6x=13/5
hay x=-13/30
Ta có: \(\left|\dfrac{1}{4}+x\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}+x=\dfrac{5}{6}\\\dfrac{1}{4}+x=\dfrac{-5}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}-\dfrac{1}{4}\\x=\dfrac{-5}{6}-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{12}-\dfrac{3}{12}\\x=\dfrac{-10}{12}-\dfrac{3}{12}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)
Vậy \(x=\dfrac{7}{12}\) hoặc \(x=-\dfrac{13}{12}\)