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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) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
\(\Leftrightarrow\left(19.75\right):x=\left(\dfrac{33}{5}-\dfrac{51}{16}\right)\cdot\dfrac{35}{6}:\dfrac{5}{2}\)
\(\Leftrightarrow19.75:x=\dfrac{637}{80}\)
hay x=1580/637
\(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right):\left(x-\dfrac{13}{7}\right)< 0\)
\(TH1:\left\{{}\begin{matrix}x-\dfrac{1}{5}>0\\x-\dfrac{13}{7}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{13}{7}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{5}< x< \dfrac{13}{7}\)
\(TH2:\left\{{}\begin{matrix}x-\dfrac{1}{5}< 0\\x-\dfrac{13}{7}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\) (vô lý nên loại)
Vậy \(\dfrac{1}{5}< x< \dfrac{13}{7}\) thỏa mãn đề bài
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}-\dfrac{1213}{100}=2\cdot\left[\left(x-\dfrac{10}{7}\right)\cdot\dfrac{49}{50}+\dfrac{2}{5}\right]\)
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}=\dfrac{49}{25}\left(x-\dfrac{10}{7}\right)+\dfrac{4}{5}+\dfrac{1213}{100}\)
\(\Leftrightarrow\dfrac{1313}{100}-\dfrac{101}{25}x=\dfrac{49}{25}x-\dfrac{490}{175}+\dfrac{1293}{100}\)
=>-6x=13/5
hay x=-13/30
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
\(a,\dfrac{x-1}{x+5}=\dfrac{6}{7}\\ \Leftrightarrow\left(x-1\right).7=6\left(x+5\right)\\ \Rightarrow7x-7=6x+30\\ \Rightarrow7x-6x=7+30\\ \Rightarrow x=37\)
Vậy \(x=37\)
\(b,\dfrac{x^2}{6}=\dfrac{24}{25}\\ \Leftrightarrow x^2.25=24.6\\ \Rightarrow x^2.5^2=144\\ \Rightarrow\left(5x\right)^2=144\\ \Rightarrow\left(5x\right)^2=\left(\pm12\right)^2\\ \Rightarrow\left\{{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Vậy \(x=\pm\dfrac{12}{5}\)
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
a/ \(\dfrac{x-1}{5}=\dfrac{1-2x}{3}\)
\(\Leftrightarrow3\left(x-1\right)=5\left(1-2x\right)\)
\(\Leftrightarrow3x-3=5-10x\)
\(\Leftrightarrow3x+10x=5+3\)
\(\Leftrightarrow13x=8\)
\(\Leftrightarrow x=\dfrac{8}{13}\)
Vậy ...
b/ \(\dfrac{3-\left|x\right|}{5}=1\dfrac{1}{2}:\dfrac{-6}{5}\)
\(\Leftrightarrow\dfrac{3-\left|x\right|}{5}=\dfrac{-5}{4}\)
\(\Leftrightarrow\left(3-\left|x\right|\right)4=5.\left(-5\right)\)
\(\Leftrightarrow\left(3-\left|x\right|\right).4=-25\)
\(\Leftrightarrow3-\left|x\right|=-6,25\)
\(\Leftrightarrow\left|x\right|=-3,25\)
\(\Leftrightarrow x\in\varnothing\)
\(\dfrac{x-1}{5}=\dfrac{1-2x}{3}\Rightarrow3x-3=5-10x\)
Áp dụng tính chất chuyển quế đổi giấu
3x+10x=5+3=8
13x=8
\(\Rightarrow\dfrac{8}{13}\)
b)\(\dfrac{3-|x|}{5}=1\dfrac{1}{2}chia\dfrac{-6}{5}=\dfrac{-5}{4}\)
3-/x/=5chia\(\dfrac{-5}{4}\)=-4
/x/=-4+3=-1
Mà /x/\(\ge0\Rightarrow x\in\varnothing\)
Tick em nha
OK you. I will help
Giải
Chia mỗi hạng tử cho BCNN (3,5,2) = 30
\(\Rightarrow\)\(2\left(x-y\right)=5\left(y+x\right)=3\left(x+z\right)=\dfrac{2\left(x-y\right)}{30}=\dfrac{5\left(y+x\right)}{30}=\dfrac{3\left(x+z\right)}{30}=\dfrac{x-y}{15}=\dfrac{y+z}{6}=\dfrac{x+z}{10}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, Ta có:
\(\dfrac{x-y}{15}=\dfrac{x+z}{10}=\dfrac{x-y-x-z}{15-10}=\dfrac{y-z}{5}\left(1\right)\)
\(\dfrac{x+z}{10}=\dfrac{y+z}{6}=\dfrac{x+z-y-z}{10-6}=\dfrac{x-y}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{y-z}{5}=\dfrac{x-y}{4}\left(đpcm\right)\)
hope you understand. Remember to brainstorm before asking questions. NHEO
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 \(x=-\dfrac{13}{12}\)