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a, 2\(xy\) - 2\(x\) + 3\(y\) = -9
(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12
2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12
(\(y-1\))(2\(x\) + 3) = -12
Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}
Lập bảng ta có:
| \(y\)-1 | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| \(y\) | -11 | -5 | -3 | -2 | -1 | 0 | 2 | 3 | 4 | 5 | 7 | 13 |
| 2\(x\)+3 | 1 | 2 | 3 | 4 | 6 | 12 | -12 | -6 | -4 | -3 | -2 | -1 |
| \(x\) | -1 | -\(\dfrac{1}{2}\) | 0 | \(\dfrac{1}{2}\) | \(\dfrac{3}{2}\) | \(\dfrac{9}{2}\) | \(-\dfrac{15}{2}\) | \(-\dfrac{9}{2}\) | -\(\dfrac{7}{2}\) | -3 | \(-\dfrac{5}{2}\) | -2 |
Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)
b, (\(x+1\))2(\(y\) - 3) = -4
Ư(4) = {-4; -2; -1; 1; 2; 4}
Lập bảng ta có:
| \(\left(x+1\right)^2\) | - 4(loại) | -2(loại) | -1(loại) | 1 | 2 | 4 |
| \(x\) | 0 | \(\pm\)\(\sqrt{2}\)(loại) | 1; -3 | |||
| \(y-3\) | 1 | 2 | 4 | -4 | -2 | -1 |
| \(y\) | -1 | 2 |
Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x;y\)) = (0; -1); (-3; 2); (1; 2)
Bài 1:
a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
a. ( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = 72
<=> 2x + 1 = 7
<=> x = 3
b. ( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = 34
<=> 2x - 1 = 3
<=> x = 2
c. ( x + 1 )3 = 2x3
<=> x + 1 = 2x
<=> x = 1
d. ( 2x + 1 )3 = 3x3
<=> 2x + 1 = 3x
<=> x = 1
( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = ( ±7 )2
<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = ( ±3 )4
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
( x + 1 )3 = ( 2x )3
<=> x + 1 = 2x
<=> x - 2x = -1
<=> -x = -1
<=> x = 1
( 2x + 1 )3 = ( 3x )3
<=> 2x + 1 = 3x
<=> 2x - 3x = -1
<=> -x = -1
<=> x = 1
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
a) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2=0+25=25\)
\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
b) \(8x^3-50x=0\)
\(2x\left(4x^2-25\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)
