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![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(x.\left(y+2\right)-y=15\)
\(\Rightarrow x.\left(y+2\right)=15+y\)
\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)
y + 2 là ước nguyên của 13
\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)
\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)
\(y+2=13\Rightarrow y=11\Rightarrow x=2\)
\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)
Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)
\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)
\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)
\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)
\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)
Vậy x = \(\frac{1}{3}\)
b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)
\(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)
\(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)
\(x.\frac{5}{6}=\frac{-2}{15}\)
\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)
Vậy x = \(\frac{-4}{25}\)
c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)
\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)
\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)
\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)
\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)
Vậy x = \(\frac{-4}{25}\)
Ủng hộ tớ nha m.n
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\frac{12}{15}=-\frac{x}{3}=-\frac{4}{y}=-\frac{z}{5}\)
\(\frac{-12}{15}=-\frac{x}{3}\)
\(\Leftrightarrow-x.15=-12.3\)
\(\Leftrightarrow-x.15=-36\)
\(\Leftrightarrow x=\frac{12}{5}\)
\(-\frac{12}{15}=-\frac{4}{y}\)
\(\Leftrightarrow-12.y=15.\left(-4\right)\)
\(\Leftrightarrow-12y=-60\)
\(\Leftrightarrow y=5\)
\(-\frac{12}{15}=\frac{z}{5}\)
\(\Leftrightarrow z.15=-12.5\)
\(\Leftrightarrow z.15=-60\)
\(\Leftrightarrow z=-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Leftrightarrow x=11\)
b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)
+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)
+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)
+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)
c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{8}{9}\)
a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)
\(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)
\(\Rightarrow x=11\)
b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)
hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)
hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)
Vậy x = 2, x = 3, x = -4
c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)
\(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)
Vậy x = 8/9
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có\(\frac{-15}{12}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\)
=>\(\frac{-15}{12}x-\frac{6}{5}x=\frac{1}{2}-\frac{3}{7}\)
=>\(\left(\frac{-15}{12}-\frac{6}{5}\right)x=\frac{1}{14}\)
=>\(\frac{-49}{20}x=\frac{1}{14}\)
=>\(x=\frac{1}{14}:\frac{-49}{20}\)
=>\(x=\frac{-10}{343}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
tui ko bít làm
theo đề ta có: x/15 = 15/x => x^2 = 15.15=15^2 =>x=15