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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{x}{-2}=\dfrac{1}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}\)
=>x=1; y=36; z=12
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a)-3/10-(-1/5)+x)=-3/2
-1/5+x =-3/10-(-3/2)
-1/5+x =6/5
x =6/5-(-1/5)
x =7/5
b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)
x-3/4+16/5 =1/2
x-3/4 =1/2-16/5
x =-27/10
x =-27/10+3/4
x =-39/20
c)x-3/x+5=4/3
=>(x-3).3=4.(x+5)
3x-9 =4x+20
3x-4x =20+9
-1x =29
x =-29
Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!
Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2
Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{-4}{15}< \dfrac{5x-1}{18}< \dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{-48}{180}< \dfrac{10\left(5x-1\right)}{180}< \dfrac{75}{180}\)
Suy ra: \(-48< 10\left(5x-1\right)< 75\)
\(\Leftrightarrow10\left(5x-1\right)\in\left\{-40;-30;-20;-10;0;10;20;30;40;50;60;70\right\}\)
\(\Leftrightarrow5x-1\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7\right\}\)
\(\Leftrightarrow5x\in\left\{-3;-2;-1;0;1;2;3;4;5;6;7;8\right\}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)(Vì x nguyên)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e: =>2/7-x=2/5
=>7-x=5
=>x=2
f: =>2x+3/3=10/3
=>2x+3=10
=>2x=7
=>x=7/2
g: =>(14+x)/7=15/7
=>x+14=15
=>x=1
h: =>(2x+3)/x=13/x
=>2x+3=13
=>2x=10
=>x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\left(ĐKXĐ:y\ne0\right)\)
\(\Rightarrow\dfrac{xy-27}{9y}=\dfrac{1}{18}\)
\(\Rightarrow18\left(xy-27\right)=9y\)
\(\Rightarrow2\left(xy-27\right)=y\)
\(\Rightarrow2xy-54=y\)
\(\Rightarrow2xy-y=54\Rightarrow y\left(2x-1\right)=54\)
\(\Rightarrow y=\dfrac{54}{2x-1}\)
- Suy ra 54 chia hết cho 2x - 1
\(\Rightarrow2x-1\inƯ\left(54\right)\)
\(\Rightarrow2x-1\in\left\{1;-1;2;-2;3;-3;9;-9;27;-27\right\}\)
Cho 2x - 1 bằng từng giá trị ở trên, ta tìm được :
\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};2;-1;5;-4;14;-13\right\}\). Mà x không có giá trị ngoài tập số nguyên.
\(\Rightarrow x\in\left\{-13;-4;-1;0;1;2;5;14\right\}\)
Thay các giá trị x trên vừa tìm được vào y :
\(\Rightarrow y\in\left\{54;-54;18;-18;6;-6;2;-2\right\}\)
Vậy : Các số x và y thỏa mãn đề bài là : \(\left(x;y\right)\in\left\{\left(1;54\right),\left(0;-54\right),\left(2;18\right),\left(-1;-18\right),\left(5;6\right),\left(-4;-6\right),\left(14;2\right),\left(-13;-2\right)\right\}\)
\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)
\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)
\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)
Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)
\(\Rightarrow x-3=0\Leftrightarrow x=3\)