\(\left|3-2x\right|-3=-\left(-3\right)\)

\(\Rightarrow\left|3-2x\right|=6\)

\(\Rightarrow\left[{}\begin{matrix}3-2x=6\\3-2x=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\).

Ta có:

|3 - 2x| - 3 = -(-3)

|3 - 2x| - 3 = 3

|3 - 2x| = 3 + 3

|3 - 2x| = 6

=> 3 - 2x = 6 hoặc 3 - 2x = -6

TH1: 3 - 2x = 6

2x = 3 - 6

2x = -3

x = -1,5

TH2: 3 - 2x = -6

2x = 3 - (-6)

2x = 3 + 6

2x = 9

x = 4,5

Vậy x = -1,5 hoặc 4,5 là giá trị cần tìm

( -7 ) - 2 ( 13 - x ) = 30

          2 ( 13 - x ) = ( -7 ) - 30

          2 ( 13 - x ) = -37

             ( 13 - x ) = -37 : 2

               13 - x   = -18,5

                      x   = 13 - ( -18,5 )

                      x   = 31,5

-7 - 2(13 - x) = 30

-2(13 - x) = 30 + 7

-2(13 - x) = 37

13 - x = 37 : (-2)

13 - x = -18,5

x = 13 - (-18,5)

x = 31,5

Dễ mà bạn :>

= 31,5 nha bạn.

\(\left(x-1\right)^5=-32\)

\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)

\(\Rightarrow x-1=-2\)

\(\Rightarrow x=-2+1\)

\(\Rightarrow x=-1\)

(x-1)⁵= -32

=>(x-1)⁵=(-2)⁵

=> x-1 = -2

=> x = -2 +1

=> x = -1.

(x-1)^5= -32

(x-1)^5= -1^5

x-1 = -1

x= -1+1

x = 0

\(sin\alpha=sin\left(180-\alpha\right)=\dfrac{3}{5}\Rightarrow cos\left(180-a\right)=\sqrt{1-sin^2\alpha}=\dfrac{4}{5}\Rightarrow cos\alpha=-\dfrac{4}{5}\)

\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\Rightarrow cot\alpha=-\dfrac{4}{3}\)

\(\Rightarrow A=\dfrac{3.\dfrac{3}{5}-\dfrac{4}{5}}{-\dfrac{3}{4}+\dfrac{4}{3}}=\dfrac{12}{7}\)