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tìm x biết :
\(3\cdot x-x\cdot y+3\cdot y=8\)
ai làm nhanh,sẽ được tích cho 1 like và nhớ là phải đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
x(3-y)+3y=8=> x(3-y)+3y-9=8-9=>x(3-y)-3(3-y)=-1=>(x-3)(3-y)=-1
Ta co bang sau
x-3 1 -1
3-y -1 1
x 4 2
y 4 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
2.x - 1 = | x - 3 |
+) TH1 :
2.x - 1 = - ( x - 3 )
2.x - 1 = -x + 3
2.x + x = 3 + 1
3.x = 4
x = 4/3
+) 2.x - 1 = x - 3
=> 2.x - x = -3 + 1
=> x = -2
Vậy x = 4/3; -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
1/2 . | 4 − 3 · x | − 2 = 1
1/2 . | 4 − 3 · x | = 1 + 2
1/2 . | 4 − 3 · x | = 3
| 4 − 3 · x | = 3 : 1/2
| 4 − 3 · x | = 6
Th 1 : 4 - 3 .x = 6
=> 3 . x = 4 - 6
[ Loại . Vì x thuộc Z ( vì lớp 6 ) ]
Th2 : 4 - 3 . x = ( - 6)
3 . x = 4 - ( - 6 )
3 . x = 4 + 6
3 . x = 10
x = 10 : 3 = 10/3
Vậy X = 10/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\cdot x-x\cdot y+3\cdot y=8\)
\(x\cdot\left(3\cdot y\right)+3\cdot y\)= 8
=>\(x\cdot\left(3-y\right)+3\cdot y-9=8-9\)
=>\(x\cdot\left(3-y\right)=-1\)
Ta có bảng sau
![](https://rs.olm.vn/images/avt/0.png?1311)
\(25\%.y+50\%.y-\frac{3}{4}.y+4.y=10\)
\(y.\left(\frac{1}{4}+\frac{1}{2}-\frac{3}{4}+4\right)=10\)
\(y.4=10\)
\(y=\frac{5}{2}\)
\(x.\frac{1}{4}-\frac{3}{4}=6:\frac{3}{4}\)
\(x.\frac{1}{4}-\frac{3}{4}=6.\frac{4}{3}\)
\(x.\frac{1}{4}=8+\frac{3}{4}\)
\(x.\frac{1}{4}=\frac{35}{4}\)
\(x=\frac{35}{4}:\frac{1}{4}\)
\(x=35\)
25% x y + 50% x y - 3/4 x y + 4 x y = 10
1/4 x y + 1/2 x y - 3/4 x y + 4 x y = 10
y x ( 1/4 + 1/2 - 3/4 + 4 ) = 10
y x 4 = 10
y = 10 : 4
y = 2.5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left|2x-3\right|\le2\)
\(\Leftrightarrow\)\(-2\le2x-3\le2\)
\(\Leftrightarrow\)\(-2+3\le2x-3+3\le2+3\)
\(\Leftrightarrow\)\(1\le2x\le5\)
\(\Leftrightarrow\)\(\frac{1}{2}\le\frac{2x}{2}\le\frac{5}{2}\)
\(\Leftrightarrow\)\(\frac{1}{2}\le x\le\frac{5}{2}\)
Vậy \(\frac{1}{2}\le x\le\frac{5}{2}\)
Chúc bạn học tốt ~