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Để | 2x + 3 | = x + 2 <=> 2x + 3 = ± ( x + 2 )
TH1 : 2x + 3 = x + 2
<=> 2x - x = 2 - 3
=> x = - 1 ( TM )
TH2 : 2x + 3 = - ( x + 2 )
<=> 2x + 3 = - x - 2
<=> 2x + x = - 2 - 3
=> 3x = - 5 ( loại )
Vậy x = - 1
![](https://rs.olm.vn/images/avt/0.png?1311)
với |2x+10|+|3x-1|+|1-x|=3 ta có 2 trường hợp:
trường hợp 1:|2x+10|+|3x-1|+|1-x|=2x+10+3x-1+1-x=3
4x+10=3
4x=-7
x=-7/4
trường hợp 2:|2x+10|+|3x-1|+|1-x|=-(2x+10)+[-(3x-1)]+[-(1-x)]=3
-2x-10-3x+1-1+x=3
-4x-10=3
-4x=13
x=-13/4
/ là dấu phần nhé!
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a) (x+2) + (x+3) + (x+5) = 25
3x + 10 = 25
3x = 15
x = 5
b) 62 - 3.(x+2) = 52.2
62 - 3.(x+2) = 50
3.(x+2) = 12
x+2 = 4
x = 2
c) 25 - (2x+3) = 16
25 - 2x - 3 = 16
22 - 2x = 16
2x =6
x = 3
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1)
xy + x - 4y = 12
x + y(x - 4) = 12
y(x - 4) = 12 - x
\(y=\dfrac{-x+12}{x-4}\)
Vì \(x,y\inℕ\) nên
\(\left(-x+12\right)⋮\left(x-4\right)\)
\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)
\(16⋮\left(x-4\right)\)
\(\left(x-4\right)\inƯ\left(16\right)\)
\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)
\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)
\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)
\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)
Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
2)
(2x + 3)(y - 2) = 15
\(\left(2x+3\right)\inƯ\left(15\right)\)
\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
Ta lập bảng
2x + 3 | 1 | -1 | 3 | -3 | 5 | -5 | 15 | -15 |
y - 2 | 15 | -15 | 5 | -5 | 3 | -3 | 1 | -1 |
(x; y) | (-1; 17) | (-2; -13) | (0; 7) | (-3; -3) | (1; 5) | (-4; -1) | (6; 3) | (-9; 1) |
Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left|2x\left(x+3\right)\right|\ge0\)
\(\Rightarrow\)\(x\ge0\)
\(PT\)\(\Leftrightarrow\)\(2x\left(x+3\right)=x\)
\(\Leftrightarrow\)\(2x^2+6x=x\)
\(\Leftrightarrow\)\(2x^2+6x-x=0\)
\(\Leftrightarrow\)\(2x^2+5x=0\)
\(\Leftrightarrow\)\(x\left(2x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-5}{2}\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=\frac{-5}{2}\)
\(\left|2-3x\right|=\left|-\frac{1}{2}\right|\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2-3x=\frac{-1}{2}\\2-3x=\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2+\frac{1}{2}\\3x=2-\frac{1}{2}\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=\frac{5}{2}\\3x=\frac{3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=\frac{5}{6}\) hoặc \(x=\frac{1}{2}\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow\hept{\begin{cases}2x-3=5\\2x-3=-5\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=-1\end{cases}}}\)
(2x-3)2=25
=>2x-3=5 hoặc 2x-3=-5
=> 2x= 8 hoặc 2x= -2
=> x=4 hoặc x=-1