a, |x-3| + x - 5 ( x < 3 )

với x < 3 , | x - 3 | + x - 5

                     = -(x-3) + x - 5 = -2

hok tốt

a) \(|x-3|+x-5\) với \(x< 3\)

\(\Rightarrow x-3+x-5\)với \(x< 3\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

Vì \(x< 3\)nên \(x=2\)

Vậy \(x=2< 3\)

Hk tốt,

k nhé

x < 3 => x - 3 < 0 => |x - 3| = -(x - 3) = 3 - x

=> M = 3 - x + x - 5 = -2

x < 3 => x - 3 < 0 => (x - 3) = - (x - 3) = 3 - x

=> M = 3 - x + x - 5 = -2

Ta có: x < 4 ⇒ | x - 4 | = 4 - x

Khi đó ta có: A = | x - 4 | - x + 1 = 4 - x - x + 1 = 5 - 2x.

Vậy A = 5 - 2x

Ta có: x < 4 ⇒ | x - 4 | = 4 - x

Khi đó ta có: A = | x - 4 | - x + 1 = 4 - x - x + 1 = 5 - 2x.

Vậy A = 5 - 2x

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

=>\(\left[\begin{array}{l}a^2-3a+1=1\\ a^2-3a+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2-3a=0\\ a^2-3a+2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}a\left(a-3\right)=0\\ \left(a-1\right)\left(a-2\right)=0\end{array}\right.\Rightarrow a\in\left\lbrace0;1;2;3\right\rbrace\)

ĐKXĐ: a<>2

=>a∈{0;1;3}

\(A=\frac{2a^3-12a^2+17a-2}{a-2}\)

\(=\frac{2a^3-4a^2-8a^2+16a+a-2}{a-2}=\frac{2a^2\left(a-2\right)-8a\left(a-2\right)+\left(a-2\right)}{a-2}\)

\(=2a^2-8a+1\)

Khi a=0 thì \(A=2a^2-8a+1=2\cdot0^2-8\cdot0+1=1\)

Khi a=1 thì \(A=2a^2-8a+1=2\cdot1^2-8\cdot1+1=2-8+1=3-8=-5\)

Khi a=3 thì \(A=2a^2-8a+1=2\cdot3^2-8\cdot3+1=18-24+1=19-24=-5\)

Với x > 0 ⇒ | 5x | = 5x

Khi đó ta có: A = 3x + 2 + | 5x | = 3x + 2 + 5x = 8x + 2

Vậy A = 8x + 2.

Ta có: x < 0 ⇒ | 4x | = - 4x

Khi đó ta có: A = | 4x | - 2x + 12 = - 4x - 2x + 12 = 12 - 6x

Vậy A = 12 - 6x.

Với x > 0 ⇒ | 5x | = 5x

Khi đó ta có: A = 3x + 2 + | 5x | = 3x + 2 + 5x = 8x + 2

Vậy A = 8x + 2.