a, a\(\ge\)0

b, a< 0

c , a \(^{ }\notinℝ\)

d a \(\le\)0

e a \(\le\)

Thời gian đi từ TT Dakmil đến TT bmt là: 11 giờ - 9 giờ 30 phút = 1 giờ 30 phút = 1,5 giờ

Độ dài quãng đường từ TT Dakmil đến TT bmt là: 40 x 1,5 = 60 ( km )

                                                               Đáp số: 60 km

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

=> \(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

=> \(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)

+) \(x+\frac{4}{15}=\frac{8}{5}\)

=> \(x=\frac{8}{5}-\frac{4}{15}=\frac{24}{15}-\frac{4}{15}=\frac{20}{15}=\frac{4}{3}\)

+) \(x+\frac{4}{15}=-\frac{8}{5}\)

=> \(x=-\frac{8}{5}-\frac{4}{15}\)

=> \(x=-\frac{24}{15}-\frac{4}{15}=-\frac{28}{15}\)

\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)

\(|x+\frac{4}{15}|-3,75=-2,15\)

\(|x+\frac{4}{15}|=-2,15+3,75\)

\(|x+\frac{4}{15}|=1,6\)

Ta có : \(|x+\frac{4}{15}|\ge0\forall x\)

\(\Rightarrow|x+\frac{4}{15}|=x+\frac{4}{15}\)

\(\Rightarrow x+\frac{4}{15}=1,6\)

\(x+\frac{4}{15}=\frac{8}{5}\)

\(x=\frac{8}{5}-\frac{4}{15}\)

\(x=\frac{4}{3}\)

a) \(\sqrt{11-6\sqrt{2}}-\sqrt{27+10\sqrt{2}}\)

\(=\sqrt{9-6\sqrt{2}+2}-\sqrt{25+10\sqrt{2}+2}\)

\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(5+\sqrt{2}\right)^2}\)

\(=\left|3-\sqrt{2}\right|-\left|5+\sqrt{2}\right|\)

\(=3-\sqrt{2}-5-\sqrt{2}=-2-2\sqrt{2}\)

b) \(\sqrt{13-4\sqrt{3}}-\sqrt{16-8\sqrt{3}}\)

\(=\sqrt{12-4\sqrt{3}+1}-\sqrt{12-8\sqrt{3}+4}\)

\(=\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}-2\right)^2}\)

\(=\left|2\sqrt{3}-1\right|-\left|2\sqrt{3}-2\right|\)

\(=2\sqrt{3}-1-2\sqrt{3}+2\)

\(=1\)

Để \(\frac{2x}{x-2}\)có giá trị nguyên thì :

\(2x⋮x+2\)

\(2x+4-4⋮x+2\)

\(2.\left(x+2\right)-4⋮x+2\)

vì \(2.\left(x+2\right)⋮x+2\)

\(\Rightarrow-4⋮x+2\)

\(\Rightarrow x+2\inƯC\left(-4\right)=\left\{-1;-2;-4;1;2;4\right\}\)

\(\Rightarrow x=\left\{-3;-4;-6;-1;0;2\right\}\)

Vậy để số hữu tỉ \(\frac{2x}{x-2}\)có giá trị nguyên thì \(x=\left\{-3;-4;-6;-1;0;2\right\}\)

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

\(=\left(1+2+3+4-3-2-1\right)+\left(-\frac{1}{2}-\frac{2}{3}-\frac{3}{4}-\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)\)

\(=4+\left[\left(-\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(-\frac{3}{4}-\frac{1}{4}\right)\right]\)

\(=4+\left[\left(-1\right)+\left(-1\right)+\left(-1\right)\right]\)

\(=4+\left(-3\right)=1\)

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

\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{2}{3}+\frac{1}{3}\right)-\left(\frac{3}{4}+\frac{1}{4}\right)\)

\(=4-1-1-1\)

\(=1\)

  11/125 - 17/18 - 5/7 + 4/9 + 17/14

=11/125 - (17/18 - 4/9) - (5/7 -17/14)

=11/125 - (17/18 - 8/18) - (10/14 - 17/14)

=11/125 - 9/18 + 7/14

=11/125 - 1/2 + 1/2

=11/125 (= 0,88)