tam giác MNK vuông ở M có NK²=MN²+MK²
                                               NK²=25+36=61
                                               NK=\(\sqrt{61}\)
sinN=\(\dfrac{MK}{NK}=\dfrac{6}{\sqrt{61}};cosN=\dfrac{MN}{NK}=\dfrac{5}{\sqrt{61}};tanN=\dfrac{6}{5};cotanN=\dfrac{5}{6}\)
\(sinK=\dfrac{5}{\sqrt{61}};cosK=\dfrac{6}{\sqrt{61}};tanK=\dfrac{5}{6};cotanK=\dfrac{6}{5}\)

\(=\left[\dfrac{3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{14\left(2\sqrt{2}+1\right)}{\left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right)}-\dfrac{4\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\right]\left(2\sqrt{2}+2\right)\)
\(=\left[\dfrac{3\sqrt{2}-3}{2-1}+\dfrac{14\left(2\sqrt{2}+1\right)}{8-1}-\dfrac{4\left(2+\sqrt{2}\right)}{4-2}\right]\left(2\sqrt{2}+2\right)\)
\(=\left[3\sqrt{2}-3+2\left(2\sqrt{2}+1\right)-2\left(2+\sqrt{2}\right)\right]\left(2\sqrt{2}+2\right)\)
\(=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
\(=5\left(\sqrt{2}-1\right).2\left(\sqrt{2}+1\right)\)
\(=5.2.\left(2-1\right)=10\)

có 
1x1=1
2x2=4
3x3=9
......
121=11x11
144=12x12
169=13x13
các số còn thiếu là
4x4=16
5x5=25
6x6=36
7x7=49
8x8=64
9x9=81
10x10=100
có
1+4+9+16+25+36+49+64+81+100+121+144+169
=819

quãng đường sài gòn - bạc liêu là 
(9x+5)(x+3)

\(A=9\sqrt{2}+7\sqrt{2}-12\sqrt{2}=4\sqrt{2}\)
\(B=\sqrt{2}-1-\sqrt{\left(3+\sqrt{2}\right)^2}=\sqrt{2}-1-3-\sqrt{2=-4}\)

uk tại bài này ngắn nên mk mới làm thế chứ bài dài thì phải nghĩ cách khác

a)
\(\dfrac{14}{10}=\dfrac{21}{15};\dfrac{15}{10}=\dfrac{21}{14};\dfrac{14}{21}=\dfrac{10}{15};\dfrac{15}{21}=\dfrac{10}{14}\)
\(\dfrac{AB}{2}=\dfrac{CD}{3};\dfrac{3}{2}=\dfrac{CD}{AB};\dfrac{2}{3}=\dfrac{AB}{CD};\dfrac{2}{AB}=\dfrac{3}{CD}\)
\(\dfrac{AB}{EF}=\dfrac{GH}{CD};\dfrac{CD}{EF}=\dfrac{GH}{AB};\dfrac{CD}{GH}=\dfrac{EF}{AB};\dfrac{AB}{GH}=\dfrac{EF}{CD}\)
\(\dfrac{4}{5}=\dfrac{AB}{MN};\dfrac{MN}{5}=\dfrac{AB}{4};\dfrac{5}{4}=\dfrac{MN}{AB};\dfrac{5}{MN}=\dfrac{4}{AB}\)

có 
1x1=1
2x2=4
3x3=9
......
121=11x11
144=12x12
169=13x13
các số còn thiếu là
4x4=16
5x5=25
6x6=36
7x7=49
8x8=64
9x9=81
10x10=100
có
1+4+9+16+25+36+49+64+81+100+121+144+169
=819

\(=2x^2-4x+5x-10-3x^2-12x-12+x^2+2x+1=-9x-21\)

a)
=10+(11+99)+(12+98)+.....+(54+56)+55
=10+55+110+110+...+110
=10+55+110.(99-11):2
=10+55+110.44
=65+4840=4905