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a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)
<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)
=> x-3=3
<=> x=6
Vậy x=6
\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)
* \(\dfrac{x}{15}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{-6}{15}\)
\(\Rightarrow x=-6\)
*\(\dfrac{4}{y}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)
\(\Rightarrow y=-10\)
Vậy x = - 6 ; y = - 10
\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)
=> ( x - 3 ) . 20 = 4. 15
=> 20x - 60 = 60
=> 20x = 60 + 60
=> 20x = 120
=> x = 120 : 20
=> x = 6
Vậy x = 6
\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)
\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)
\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)
\(\Rightarrow-3+\left(-1\right)< x\le-1\)
\(\Rightarrow-4< x\le-1\)
\(\Rightarrow x=-3;-2;-1\)
2. \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-3}{2}\right)=\dfrac{-21}{4}.\dfrac{2}{7}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-15}{10}\right)=\dfrac{-3}{2}\)
\(\Leftrightarrow x.\dfrac{6}{5}=\dfrac{-3}{2}\)
\(\Leftrightarrow x=\dfrac{-3}{2}:\dfrac{6}{5}\)
\(\Leftrightarrow x=\dfrac{-3}{2}.\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{-5}{4}\)
3.\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=1\\2x-\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1+\dfrac{3}{4}\\2x=\left(-1\right)+\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=\dfrac{-7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}.\dfrac{1}{2}\\x=\dfrac{-7}{3}.\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{7}{6};\dfrac{-7}{6}\right\}\)
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
a) Ta có: \(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)
\(\Rightarrow\left(5-2y\right)x=24\)
Vì \(x,y\in Z\Rightarrow\left[{}\begin{matrix}5-2y\in Z\\x\in Z\end{matrix}\right.\)
\(\Rightarrow5-2y\inƯ\left(24\right);x\inƯ\left(24\right)\)
Tự lập bảng xét các giá trị của \(x,y\) nhé.
b) Lại có: \(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1+2y}{6}\)
\(\Rightarrow\left(1+2y\right)x=30\)
Lí luận rồi lập bảng như câu \(a\)).
c) \(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\)
\(\Rightarrow\dfrac{2}{y}=\dfrac{x}{6}-\dfrac{1}{30}\)
\(\Rightarrow\dfrac{2}{y}=\dfrac{5x-1}{30}\)
\(\Rightarrow\left(5x-1\right)y=60\)
\(......Tương\) \(tự\) \(như\) \(câu\) \(a\))\(b\)).
a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)
⇒\(2x+1=21\)
\(2x=21-1\)
\(2x=20\)
⇒\(x=10\)