x=-36

x=11

x=100

x=14

x=-36

x=11

x=100

x=14

2: \(\Leftrightarrow\left(\dfrac{x+1}{12}-1\right)+\left(\dfrac{x+2}{13}-1\right)=\left(\dfrac{x+3}{14}-1\right)+\left(\dfrac{x+4}{15}-1\right)\)

=>x-11=0

=>x=11

3: \(\Leftrightarrow\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

=>x-100=0

=>x=100

a) x(x - 5) - 4x + 20 = 0

\(\Leftrightarrow\) x(x - 5) - (4x + 20)

\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0

\(\Leftrightarrow\) (x - 5)(x - 4)

Khi x - 5 = 0 hoặc x - 4 = 0

 \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 4

 Vậy S = \(\left\{5;4\right\}\)

b) x(x + 6) - 7x - 42 = 0

 \(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0

 \(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0

 \(\Leftrightarrow\) (x + 6)(x - 7) = 0

Khi x - 6 = 0 hoặc x - 7 = 0

   \(\Leftrightarrow\) x = 6           \(\Leftrightarrow\) x = 7

 Vậy S = \(\left\{6;7\right\}\)

c) x³ - 5x² - x + 5 = 0

 \(\Leftrightarrow\) (x³ - 5x²) - (x + 5) = 0

 \(\Leftrightarrow\) x² (x - 5) - (x - 5) = 0

 \(\Leftrightarrow\) (x - 5)(x² - 1) = 0

 \(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0

 Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

   \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 1            \(\Leftrightarrow\) x = -1

 Vậy S = \(\left\{5;1;-1\right\}\)

d) 4x² - 25 - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x)² - 5² - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0

\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0

Khi 2x - 5 = 0 hoặc -x + 12 = 0

  \(\Leftrightarrow\) 2x = 5             \(\Leftrightarrow\)   -x = -12

  \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)              \(\Leftrightarrow\) x = 12

 Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)

e) x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) x³ - 3³ + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0

\(\Leftrightarrow\) (x - 3) ( x² - 3x + 9 + x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 2x) = 0

\(\Leftrightarrow\) (x - 3)x(x - 2)

 Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0

    \(\Leftrightarrow\) x = 3                            \(\Leftrightarrow\) x = 2

 Vậy S = \(\left\{3;0;2\right\}\)

 Chúc bạn học tốt

a) Ta có: \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

b) Ta có: \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

c) Ta có: \(x^3-5x^2-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)

d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

a) \(x+15=18-\left(-27\right)\)

\(\Leftrightarrow x+15=18+27\)

\(\Leftrightarrow x+15=45\)

\(\Leftrightarrow x=45-15\)

\(\Leftrightarrow x=30\)

b) \(-x+7=28\)

\(\Leftrightarrow-x=28-7\)

\(\Leftrightarrow-x=21\)

\(\Leftrightarrow x=-21\)

c) \(20-\left(-x\right)=14\)

\(\Leftrightarrow20+x=14\)

\(\Leftrightarrow x=14-20\)

\(\Leftrightarrow x=-6\)

d) \(2x+5=x-14\)

\(\Leftrightarrow2x+5-\left(x-14\right)=0\)

\(\Leftrightarrow2x+5-x+14=0\)

\(\Leftrightarrow x+19=0\)

\(\Leftrightarrow x=0-19\)

\(\Leftrightarrow x=-19\)

e) \(x+17=-8-20\)

\(\Leftrightarrow x+17=-\left(8+20\right)\)

\(\Leftrightarrow x+17=-28\)

\(\Leftrightarrow x=-28-17\)

\(\Leftrightarrow x=-\left(28+17\right)\)

\(\Leftrightarrow x=-45\)

f) \(3x+15=2x+6\)

\(\Leftrightarrow3x+15-\left(2x+6\right)=0\)

\(\Leftrightarrow3x+15-2x-6=0\)

\(\Leftrightarrow x+9=0\)

\(\Leftrightarrow x=0-9\)

\(\Leftrightarrow x=-9\)

2,

a.\(\left|a+3\right|=\left|-5\right|\)

\(a+3=5\)

\(a=2\)

Vậy a=2

b.\(\left|a-1\right|=2+\left|-3\right|\)

\(a-1=2+3\)

\(a-1=5\)

\(a=6\)

Vậy a=6

1/

\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)

\(\Leftrightarrow43x=0\Rightarrow x=0\)

2/

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)

\(\Leftrightarrow x^3+27-x^3+x=27\)

\(\Leftrightarrow x=0\)

a)      Chỗ sai trong phương trình là: \(5 - x + 8 = 3x + 3x - 27\) (dòng thứ 2) vì khi phá ngoặc đã không đổi dấu của số 8.

Sửa lại:

\(\begin{array}{l}5 - \left( {x + 8} \right) = 3x + 3\left( {x - 9} \right)\\\,\,\,\,5 - x - 8 = 3x + 3x - 27\\\,\,\,\,\,\,\, - 3 - x = 6x - 27\\\,\,\,\, - x - 6x =  - 27 + 3\\\,\,\,\,\,\,\,\,\,\,\,\, - 7x =  - 24\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \left( { - 24} \right):\left( { - 7} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{24}}{7}\end{array}\)

Vậy phương trình có nghiệm \(x = \frac{{24}}{7}.\)

b)     Chỗ sai trong phương trình là: \(4x + 5x = 9 - 18\) (dòng thứ 3) vì khi chuyển \( - 18\) từ vế trái sang vế phải đã không đổi dấu thành \( + 18\).

Sửa lại:

\(\begin{array}{l}3x - 18 + x = 12 - \left( {5x + 3} \right)\\\,\,\,\,\,\,\,4x - 18 = 12 - 5x - 3\\\,\,\,\,\,\,\,4x + 5x = 9 + 18\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x = 27\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 27:9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3.\end{array}\)

Vậy phương trình có nghiệm \(x = 3.\)