Question

Tìm x € Z biết: 

a) x - (11 - x) = -48 + (-12 + x)

b) (15 - x) + (x - 12) = 7 - (-8 + x)

c) (x - 12) - (2x + 31) = -6 - 5

d) |x + 5| - (-17) = 20   ;    12x² - 3x = 0

Answer 1

a) x - (11 - x) = -48 + (-12 + x)

x - 11 + x = -48 + (-12) + x

x + x - x = -48 + (-12) + 11

x + 0 = (-60) + 11

x = -49

b) (15 - x) + (x - 12) = 7 - (-8 + x)

15 - x + x - 12 = 7 + 8 + x

15 - x + x - 12 = 15 + x

x + x - x = 15 - 15 + 12

x + 0 = 0 + 12

x = 12

c) (x - 12) - (2x + 31) = -6 - 5

x - 12 - 2x - 31 = -1

x - 12 - 2x = -1 + 31

x - 12 - 2x = 30

x - 2x = 30 + 12

-1x = 42

=> x = -42

d) |x + 5| - (-17) = 20

=> |x + 5| + 17 = 20

=> |x + 5| = 20 - 17

=> |x + 5| = 3

=> \(\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)

=> \(\orbr{\begin{cases}x=3-5\\x=-3-5\end{cases}}\)

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

12x² - 3x = 0

=> 12x² - 3x = (2² . 3x²) - 3x = 0

=> 3x . (4x - 1) = 0

=> 3x . 4x - 3x . 1 = 0

=> 3x . 4x - 3x = 0

=> 12x - 3x = 0

=> 4x  =0

=> x = 0

Answer 2

x=0 chac chan

Answer 3

Thanks bạn nhìu!

Answer 4

câu d phần 2 bạn ấy làm sai rồi

\(12x^2-3x=0\)

\(\Leftrightarrow\)3x.(4x-1)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=0\\4x-1=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)