Tìm số nguyên x biết : x-\(\dfrac{2}{15}\)=\(\dfrac{-7}{12}\)
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Bài 4:
a) \(\dfrac{2.7.13}{26.35}=\dfrac{2.7.13}{13.2.7.5}=\dfrac{1}{5}\)
b) \(\dfrac{23.5-23}{4-27}=\dfrac{23.\left(5-1\right)}{-23}=\dfrac{23.4}{-23}=-4\)
c) \(\dfrac{2130-15}{3550-25}=\dfrac{2115}{3525}=\dfrac{3}{5}\)
Tìm số nguyên x, biết:
a) \(\dfrac{-28}{35}=\dfrac{16}{x};\) b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}.\)
\(a.\)
\(\dfrac{-28}{35}=\dfrac{16}{x}\)
\(\Rightarrow x=\dfrac{35\cdot16}{-28}=\dfrac{5\cdot7\cdot4\cdot4}{-7\cdot4}=-20\)
\(b.\)
\(\dfrac{x+7}{15}=\dfrac{-24}{36}\)
\(\Rightarrow x+7=\dfrac{15\cdot-24}{36}=\dfrac{5\cdot3\cdot-12\cdot2}{12\cdot3}=-10\)
\(\Leftrightarrow x=-17\)
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
\(\dfrac{-5}{6}\)- x=\(\dfrac{7}{12}\)+\(\dfrac{-8}{12}\)=\(\dfrac{-15}{12}\)
\(\Rightarrow\)x=\(\dfrac{-5}{6}\)-\(\dfrac{-15}{12}\)=\(\dfrac{-10}{12}\)-\(\dfrac{-15}{12}\)=\(\dfrac{-25}{12}\)
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
ĐKXĐ:\(x-1\ne0\Rightarrow x\ne1\)
\(\dfrac{4}{x-1}=\dfrac{3}{15}\\ \Leftrightarrow\dfrac{4}{x-1}=\dfrac{1}{5}\\ \Leftrightarrow x-1=5.4\\ \Leftrightarrow x-1=20\\ \Leftrightarrow x=21\)
⇔x=-9/20=-0,45
\(x-\dfrac{2}{15}=-\dfrac{7}{12}\)
⇔\(x=-\dfrac{9}{20}\)