Ta có : |x - 2| ; |x - 5| ; |x - 18| ≥0∀x∈R≥0∀x∈R

=> |x - 2| + |x - 5| + |x - 18|  ≥0∀x∈R≥0∀x∈R

=> D có giá trị nhỏ nhất khi x = 2;5;18

Mà x ko thể đồng thời nhận 3 giá trị

Nên GTNN của D là : 16 khi x = 5   ok nha bạn

x^2/x-1 = x^2-4x+4/x-1 + 4 = (x-2)^1/x-1 + 4 >= 4

Dấu "=" xảy ra <=> x-2 = 0 <=> x = 2 (tm)

Vậy GTNN của x^2/x-1 = 4 <=> x= 2

k mk nha

\(2x^3-8x^2-6x+36\)

=\(2\left(x^3-4x^2-3x+18\right)\)

=\(2\left(x^3-6x^2+9x+2x^2-12x+18\right)\)

=\(2\left[x\left(x^2-6x+9\right)+2\left(x^2-6x+9\right)\right]\)

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

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

HT

\(\frac{2x}{x^2-3x}+\frac{2x}{x^2-4x+3}+\frac{x}{x-1}\)

\(=\frac{2x}{x\left(x-3\right)}+\frac{2x}{x^2-3x-x+3}+\frac{x}{x-1}\)

\(=\frac{2}{x-3}+\frac{2x}{x\left(x-3\right)-\left(x-3\right)}+\frac{x}{x-1}\)

\(=\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{2x}{\left(x-3\right)\left(x-1\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}\)

\(=\frac{2x-2+2x+x^2-3x}{\left(x-3\right)\left(x-1\right)}\)

\(=\frac{x^2+x-2}{\left(x-3\right)\left(x-1\right)}=\frac{x^2-x+2x-2}{\left(x-3\right)\left(x-1\right)}=\frac{x\left(x-1\right)+2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\frac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(x-1\right)}=\frac{x+2}{x-3}\)

5x^3/y^2 : x^2/y^3=5x^3/y^2 * y^3/x^2=5x^3*y^3/y^2*x^2=5xy