a, -x+28 = 34

=> -x = 34-28 =6

=> x= -6

b, -57-(11 - x ) =8

=> 11 - x = -57-8 = -65

=> x= 11 -(-65) = 76

c, 30-|-x - 30 | =0

=> |-x - 30 |= 30

=>\(\orbr{\begin{cases}-x-30=30\\-x-30=-30\end{cases}}\)

 <=>\(\orbr{\begin{cases}-x=30+30=60\\-x=-30=30=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=-60\\x=0\end{cases}}\)

mk thử lại rồi nha bn đúng đó nha :))

a,-x+28=34

        -x=34-28

        -x=6

=>x= -6

Vậy x= -6

b,-57 -(11-x) =8

          11-x  = -57-8

          11-x  = -57+ (-8)

          11-x  = -65

              x  = 11- (-65)

              x  = 11+65

              x  = 76

Vậy x= 76

c, 30 -|-x - 30| = 0

         |-x - 30| = 30-0

         |-x - 30| =30

Vì |30| = |-30| =30

=> -x - 30 (- { -30 ; 30 }

* -x - 30 = -30

         -x = -30 + 30

         -x = 0

=> x= 0

* -x - 30 = 30

         -x = 30 + 30

         -x = 60

=> x = -60

Thử lại : Đúng 

Vậy x (- { 0 ; -60 }

ai tra loi nhanh cai

-2 tick nhé Tran Thi Xuan

x-3/2<0

=>x<3/2

Vậy x<3/2

Vì \(\begin{matrix}\left|2,5-x\right|\ge0\forall x\\\left|x-3\right|\ge0\forall x\end{matrix}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|2,5-x\right|=0\\\left|x-3\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2,5-x=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2,5\\x=3\end{matrix}\right.\) ( vô lí )

Vậy ko có giá trị nào thỏa mãn yêu cầu đề bài .

Do |2,5-x|\(\ge0\forall x\)

|x-3|\(\ge0\forall x\)

=>\(\left|2,5-x\right|+\left|x-3\right|=0\)

=>\(\left[{}\begin{matrix}\left|2,5-x\right|=0\\\left|x-3\right|=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2,5-x=0\Rightarrow x=2,5\\x-3=0\Rightarrow x=3\end{matrix}\right.\)

vậy x=2,5 hoặc x=3

\(\left(x-1\right)^2=\left(x-3\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-3\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left[\left(x-3\right)^2\right]^2=0\)

\(\Leftrightarrow\left[\left(x-1\right)-\left(x-3\right)^2\right]\left[\left(x-1\right)+\left(x-3\right)^2\right]=0\)

\(\Leftrightarrow\left(x-1-x^2+6x-9\right)\left(x-1+x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(-x^2+7x-10\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)

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

Vậy: ... 

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