1. x² - 5x = 0
<=> x(x - 5) = 0
<=> \(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)(thỏa mãn)
2. |x - 9|.(-8) = -16
<=> |x - 9| = 2
<=> \(\left[{}\begin{matrix}x-9=2\\x-9=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=7\end{matrix}\right.\)(thỏa mãn)
@Lương Thị Quỳnh Trang

9. x - (-25 + 7) > 12 - (15 - 14)
<=> x + 18 > 11
<=> x > -7
10. |17 + (x - 15)| < 5
<=> |2 + x| < 5
<=> \(\left[{}\begin{matrix}2+x< 5\\-2-x< 5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>-7\end{matrix}\right.\)<=> -7 < x < -3
=> x = -6; -5; -4
@Lương Thị Quỳnh Trang

7. 35 - |2x - 1| = 14
<=> |2x - 1| = 21
<=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)(thỏa mãn)
8. |3x - 2| + 5 = 9 - x
<=> |3x - 2| = 4 - x
<=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=-4+x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\)(thỏa mãn)
@Lương Thị Quỳnh Trang

5. 13 - |x| = |-4|
<=> 13 - |x| = 4
<=> |x| = 9
<=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
6. |x| - 3 + 6 = 16
<=> |x| + 3 = 16
<=> |x| = 13
<=> \(\left[{}\begin{matrix}x=13\\x=-13\end{matrix}\right.\)
@Lương Thị Quỳnh Trang

3. -(x - 6 + 85) = (x + 51) - 54
<=> -x + 6 - 85 = x + 51 - 54
<=> -x - 79 = x -3
<=> -79 + 3 = x + x
<=> -76 = 2x
<=> x = -38
4. -(35 - x) - (37 - x) = 33 - x
<=> -35 + x - 37 + x = 33 - x
<=> -72 + 2x = 33 - x
<=> 2x + x = 33 + 72
<=> 3x = 105
<=> x = 35
@Lương Thị Quỳnh Trang

siêu nhân có khác

xy + 2x + y = 50
<=> xy + 2x + y + 2 = 52
<=> x(y + 2) + (y + 2) = 52
<=> (x + 1)(y + 2) = 52
Do x; y \(\in Z\Rightarrow x+1;y+2\in Z\)
Mà (x + 1)(y + 2) = 52
=> x + 1; y + 2 \(\inƯ\left(52\right)=\left\{\pm1;\pm2;\pm4;\pm13;\pm26;\pm52\right\}\)
Ta có bảng :

@Thị Hà Anh Nguyễn

1. Do (x - 7)(x + 3) < 0
=> x - 7; x + 3 khác dấu

Mặt khác x - 7 < x + 3
=> \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)<=> -3 < x < 7
Vậy x = -2; -1; 0; 1; 2; 3; 4; 5; 6
@pham thu hoai

a, Do (x - 2)(5 - x) > 0
=> x - 2; 5 - x cùng dấu
Nếu \(\left\{{}\begin{matrix}x-2>0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< 5\end{matrix}\right.\)<=> 2 < x < 5
Nếu \(\left\{{}\begin{matrix}x-2< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>5\end{matrix}\right.\)(vô lý)
Vậy x = 3; 4
b, Do (x - 3)(x - 7) < 0
=> x - 3; x - 7 khác dấu
Nếu \(\left\{{}\begin{matrix}x-3>0\\x-7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\x< 7\end{matrix}\right.\)<=> 3 < x < 7
Nếu \(\left\{{}\begin{matrix}x-3< 0\\x-7>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x>7\end{matrix}\right.\)(vô lý)
Vậy x = 4; 5; 6
@Vũ Việt Anh

5n + 7 \(⋮3n+2\)
Có 3(5n + 7) \(⋮3n+2\)
Và 5(3n + 2) \(⋮3n+2\)
=> 15n + 7 - 15n - 2 \(⋮3n+2\)
=> 5 \(⋮3n+2\)
=> 3n + 2 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
=> 3n = -1; -3; 3; -7
=> n = -\(\dfrac{1}{3};-1;1;-\dfrac{7}{3}\) (thỏa mãn)
@Selina Moon