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a) -2x -12 = 8 - 6x + 60
\(\Leftrightarrow\)-2x + 6x = 8 + 60 + 12
\(\Leftrightarrow\)4x = 80
\(\Leftrightarrow\)x= 80 : 4
\(\Leftrightarrow\)x=20
vậy x = 20
b) -8x - 36 = ( x+13 ) + ( -8x + 3 )
-8x - 36 = x + 13 - 8x + 3
-8x -x + 8x = 13 + 3 + 36
-x = 52
x = - 52
Vậy x = - 52
c) 1 + x - 10 - 6x = 4 - 5x
x - 6x + 5x = 4 -1 + 10
0x = 13
vậy x = 0
d)6 - 3x + 1 = -3x + 7
-3x + 3x = 7 - 6 - 1
0x = 0
vậy x = 0
Trần Tuyết Tâm
a) <=> -2x+6x=8+60+12
<=> 4x = 80
<=> x= 20
b) <=> -8x-36=x+13-8x+3
<=> -8x+8x-x=13+3+36
<=> -x=52 => x=-52
c) <=> x+5x-6x=4-1+10
<=> 0x = 13 => x=0
d) <=> -3x+3x=7-1-6
<=> 0x=0 => x=0
XL mik ko ghi lại đề thông cảm
a) Để \(-1:x\)là số nguyên
\(\Rightarrow\)\(x\inƯ\left(-1\right)\in\left\{\pm1\right\}\)
Vậy \(x\in\left\{-1;1\right\}\)
b) Để \(1:x+1\)là số nguyên
\(\Rightarrow\)\(x+1\inƯ\left(1\right)\in\left\{\pm1\right\}\)
+ \(x+1=1\)\(\Leftrightarrow\)\(x=1-1=0 \left(TM\right)\)
+ \(x+1=-1\)\(\Leftrightarrow\)\(x=-1-1=-2\left(TM\right)\)
Vậy \(x\in\left\{-2; 0\right\}\)
c) Để \(-2:x\)là số nguyên
\(\Rightarrow\)\(x\inƯ\left(-2\right)\in\left\{\pm1;\pm2\right\}\)
Vậy \(x\in\left\{-1;-2;1;2\right\}\)
d) Để \(3:x-2\)là số nguyên
\(\Rightarrow\)\(x-2\inƯ\left(3\right)\in\left\{\pm1;\pm3\right\}\)
- Ta có bảng giá trị:
| \(x-2\) | \(-1\) | \(1\) | \(-3\) | \(3\) |
| \(x\) | \(1\) | \(3\) | \(-1\) | \(5\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-1;1;3;5\right\}\)
e) Ta có: \(x+8=\left(x-7\right)+15\)
- Để \(x+8⋮x-7\)\(\Leftrightarrow\)\(\left(x-7\right)+15⋮x-7\)mà \(x-7⋮x-7\)
\(\Rightarrow\)\(15⋮x-7\)\(\Rightarrow\)\(x-7\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
- Ta có bảng giá trị:
| \(x-7\) | \(-1\) | \(1\) | \(-3\) | \(3\) | \(-5\) | \(5\) | \(-15\) | \(15\) |
| \(x\) | \(6\) | \(8\) | \(4\) | \(10\) | \(2\) | \(12\) | \(-8\) | \(22\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-8;2;4;6;8;10;12;22\right\}\)
f) Ta có: \(2x+9=\left(2x-10\right)+19=2.\left(x-5\right)+19\)
- Để \(2x+9⋮x-5\)\(\Leftrightarrow\)\(2.\left(x-5\right)+19⋮x-5\)mà \(2.\left(x-5\right)⋮x-5\)
\(\Rightarrow\)\(19⋮x-5\)\(\Rightarrow\)\(x-5\inƯ\left(19\right)\in\left\{\pm1;\pm19\right\}\)
- Ta có bảng giá trị:
| \(x-5\) | \(-1\) | \(1\) | \(-19\) | \(19\) |
| \(x\) | \(4\) | \(6\) | \(-14\) | \(24\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-14;4;6;24\right\}\)
g) Ta có: \(2x+16=\left(2x-16\right)+32=2.\left(x-8\right)+32\)
- Để \(2x+16⋮x-8\)\(\Leftrightarrow\)\(2.\left(x-8\right)+32⋮x-8\)mà \(2.\left(x-8\right)⋮x-8\)
\(\Rightarrow\)\(32⋮x-8\)\(\Rightarrow\)\(x-8\inƯ\left(32\right)\in\left\{\pm1;\pm2;\pm4;\pm8;\pm16;\pm32\right\}\)
- Ta có bảng giá trị:
| \(x-8\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-4\) | \(4\) | \(-8\) | \(8\) | \(-16\) | \(16\) | \(-32\) | \(32\) |
| \(x\) | \(7\) | \(9\) | \(6\) | \(10\) | \(4\) | \(12\) | \(0\) | \(16\) | \(-8\) | \(24\) | \(-24\) | \(40\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-24;-8;0;4;6;7;9;10;12;16;24;40\right\}\)
h) Ta có: \(5x+2=\left(5x-5\right)+7=5.\left(x-1\right)+7\)
- Để \(5x+2⋮x-1\)\(\Leftrightarrow\)\(5.\left(x-1\right)+7⋮x-1\)mà \(5.\left(x-1\right)⋮x-1\)
\(\Rightarrow\)\(7⋮x-1\)\(\Rightarrow\)\(x-1\inƯ\left(7\right)\in\left\{\pm1;\pm7\right\}\)
- Ta có bảng giá trị:
| \(x-1\) | \(-1\) | \(1\) | \(-7\) | \(7\) |
| \(x\) | \(0\) | \(2\) | \(-6\) | \(8\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-6;0;2;8\right\}\)
k) Ta có: \(3x=\left(3x-6\right)+6=3.\left(x-2\right)+6\)
- Để \(3x⋮x-2\)\(\Leftrightarrow\)\(3.\left(x-2\right)+6⋮x-2\)mà \(3.\left(x-2\right)⋮x-2\)
\(\Rightarrow\)\(6⋮x-2\)\(\Rightarrow\)\(x-2\inƯ\left(6\right)\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
- Ta có bảng giá trị:
| \(x-2\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(-3\) | \(3\) | \(-6\) | \(6\) |
| \(x\) | \(1\) | \(3\) | \(0\) | \(4\) | \(-1\) | \(5\) | \(-4\) | \(8\) |
| \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(x\in\left\{-4;-1;0;1;3;4;5;8\right\}\)
c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)
hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)
f: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
a) \(\frac{3}{2}-\left(x-\frac{7}{3}\right)=\left|-\frac{3}{4}-\frac{9}{8}\right|\)
=> \(\frac{3}{2}-x+\frac{7}{3}=\left|-\frac{15}{8}\right|\)
=> \(\frac{3}{2}-x+\frac{7}{3}=\frac{15}{8}\)
=> \(\frac{3}{2}-x=-\frac{11}{24}\)
=> \(x=\frac{47}{24}\)
b) \(\frac{5}{2}-\left(\frac{3}{2}-\frac{7}{3}+x\right)=\frac{8}{15}-\left(\frac{1}{4}-\frac{7}{10}\right)\)
=> \(\frac{5}{2}-\frac{3}{2}+\frac{7}{3}-x=\frac{8}{15}-\left(-\frac{9}{20}\right)\)
=> \(\frac{10}{3}-x=\frac{59}{60}\)
=> \(x=\frac{10}{3}-\frac{59}{60}=\frac{47}{20}\)
c) \(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
=> \(\frac{3}{2}-10x-\frac{4}{5}+3x=0\)
=> \(\left(\frac{3}{2}-\frac{4}{5}\right)+\left(-10x+3x\right)=0\)
=> \(\frac{7}{10}-7x=0\)
=> \(7x=\frac{7}{10}\)
=> x = 1/10
1) 3x - 6 = 5x + 2
=> 3x - 5x = 2 + 6
=> -2x = 8
=> x = -4
2) 15 - x = 4x - 5
=> 15 + 5 = 4x + x
=> 20 = 5x
=> x = 4
3) x - 15 = 6 + 4x
=> x - 4x = 6 + 15
=> -3x = 21
=> x = -7
4) -12 + x = 5x - 20
=> x - 5x = -20 + 12
=> -4x = -8
=> x = 2
5) 7x - 4 = 20 + 3x
=> 7x - 3x = 20 + 4
=> 4x = 24
=> x = 6
1) 3x- 6 = 5x + 2
5x - 3x = -6 - 2
2x = -8 => x = -4
Tương tự như trên
bt=-5(x-3y)^2-2(x-3y)-4
thay x-3y=-7 ta có:
-5*(-7)^2-2*(-7)-4=-245+14-4=-235
a) Ta có: -5(x - 3y)2 - 2x + 6y - 4
= -5.(-7)2 - 2(x - 3y) - 4
= -5 . 49 - 2.(-7) - 4
= -245 + 14 - 4
= -235
b) *, -2.(x + 6) = 8 - 6.(x - 10)
=> -2x - 12 = 8 - 6x + 60
=> -2x + 6x = 68 + 12
=> 4x = 80
=> x = 80 : 4
=> x = 20
* ,2 tương tự
a, 1+x-10-6x=4-5x
<=> -5x-9=4-5x
<=>0x=13(vô lý)
vậy phương trình vô nghiệm
b, 6-3x+1=-3x+7
-3x+3x=7-7
<=>0x=0(luôn đúng)
vậy phương trình có vô số nghiệm