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\(\left(3x-1\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(3x+3-4\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(-4\right)⋮\left(x+1\right)\)
\(\Rightarrow x+1\inƯ\left(-4\right)=\left\{-4;-1;1;4\right\}\)
\(\Rightarrow x\in\left\{-5;-2;0;3\right\}\)
a) \(\frac{5}{6}=\frac{x-1}{x}\)
\(5x=6x-6\)
\(6x-5x=6\)
\(x=6\)
các câu còn lại lm tương tự
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b) \(\frac{1}{2}=\frac{x+1}{3x}\)
\(\Rightarrow1.3x=2.\left(x+1\right)\)
\(3x=2x+2\)
\(3x-2x=2\)
\(x=2\)
Vậy x=2
các câu khác bạn làm tương tự
Câu a:
- \(\frac12\)(3 - 2\(x\)) - 7 = 5 - \(\frac13\)(\(x\) - \(\frac45\))
- \(\frac32\) + \(x\) - 7 = 5 - \(\frac13x\) + \(\frac{4}{15}\)
\(x\) + \(\frac13x\) = 5 + \(\frac{4}{15}\) + 7 + \(\frac32\)
(1 + \(\frac13\))\(x\) = \(\frac{150}{30}\) + \(\frac{8}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{158}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{368}{30}\) + \(\frac{45}{30}\)
\(\frac43x\) = \(\frac{413}{30}\)
\(x\) = \(\frac{413}{30}\) : \(\frac43\)
\(x\) = \(\frac{413}{40}\)
Vậy \(x=\frac{413}{40}\)
Câu b:
(5 - 3x/2) : - 1 3/8 = - 7 1/3
(5 - 3x/2) : (-11/8) = - 22/3
5 - 3x/2 = - 22/3 x (-11/8)
5 - 3x/2 = 121/12
3x/2 = 5 - 121/12
3x/2 = - 61/12
x = - 61/12 : 3/2
x = -61/18
Vậy x = - 61/18
a) (3x - 1)2 = 100
(3x - 1)2 = 102
=>3x - 1 = 10
=> 3x = 10 + 1
3x = 11
x = 11/3
Câu a:
\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)
-8.(-7) = 4.(3\(x\) - 1)
56 = 12\(x\) - 4
12\(x\) = 56+ 4
12\(x\) = 60
\(x\) = 60 : 12
\(x\) = 5
Vậy \(x\) = 5
Câu b:
\(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(x^2\) = (-3)\(^2\)
\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)
Vậy \(x\in\left\lbrace-3;3\right\rbrace\)
Câu c:
\(-\frac{4}{y}=\frac{x}{2}\)
-4.2 = \(x.y\)
\(xy=-8\)
Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}
Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)
Câu 2:
(\(x-1)\)(y + 2) = 7
Ư(7) = {-7; -1; 1; 7}
Lập bảng ta có:
\(x\)-1 | -7 | -1 | 1 | 7 |
\(x\) | -6 | 0 | 2 | 8 |
y+2 | -1 | -7 | 7 | 1 |
y | -3 | -9 | 5 | -1 |
\(x;y\in Z\) | tm | tm | tm | tm |
Theo bảng trên ta có:
(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)
Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)
a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
7) ( 2x2+ 1)3= 729
=> ( 2x2+ 1)3 = 36
=> ( 2x2+ 1)3= 93
=> 2x2+ 1= 9
=> 2x2= 8
=> x2= 4
=> x= 2
8) ( 3x2- 43)3= 125
=> (3x2- 43)3= 53
=> 3x2- 43= 5
=> 3x2= 48
=> x2= 16
=> x= 4
Vậy x=4