Bài 1: 

a) 4√24 - 3√54 + 5√6 - √150

= 8√6 - 9√6 + 5√6 - 5√6

= -√6

a: \(=4\cdot2\sqrt{6}-3\cdot3\sqrt{6}+5\sqrt{6}-5\sqrt{6}\)

\(=8\sqrt{6}-9\sqrt{6}=-\sqrt{6}\)

b: \(=2\left(3-2\sqrt{2}\right)-2\left(3+2\sqrt{2}\right)\)

\(=6-4\sqrt{2}-6-4\sqrt{2}=-8\sqrt{2}\)

c: \(=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)

\(\dfrac{5}{12}+\dfrac{3}{16}=\dfrac{5\cdot4+3\cdot3}{48}=\dfrac{29}{48}\)

\(\dfrac{4}{15}-\dfrac{2}{9}=\dfrac{4\cdot3-2\cdot5}{45}=\dfrac{2}{45}\)

a) \(\dfrac{5}{12}+\dfrac{3}{16}=\dfrac{20}{48}+\dfrac{9}{48}=\dfrac{29}{48}\)

b) \(\dfrac{4}{15}-\dfrac{2}{9}=\dfrac{12}{45}-\dfrac{10}{45}=\dfrac{2}{45}\)

\(a,\dfrac{5}{12}+\dfrac{3}{16}=\dfrac{20}{48}+\dfrac{9}{48}=\dfrac{29}{48}\)

\(b,\dfrac{4}{15}-\dfrac{2}{9}=\dfrac{12}{45}-\dfrac{10}{45}=\dfrac{12-10}{45}=\dfrac{2}{45}\)

\(A=\dfrac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(=\dfrac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)

\(=\dfrac{5.2^{20}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}\left(5-7.2\right)}\)

\(=\dfrac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-14\right)}=\dfrac{2.9}{-9}=-2\)

AI LÀM ĐƯỢC TẶNG 2  COIN NÈ MẠI DÔ

khó:)))

a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)

\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)

b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)

\(\dfrac{3}{5}+\dfrac{13}{25}-\dfrac{5}{9}-\dfrac{3}{7}\)

\(=\dfrac{15}{25}+\dfrac{13}{25}-\dfrac{35}{63}-\dfrac{27}{63}\)

\(=\dfrac{28}{25}-\dfrac{62}{63}=\dfrac{214}{1575}\)

bằng 214/1575

a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)

\(=127-22\sqrt{6}\)

b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

=-1+5

=4

1

a: \(-15-\left(-13+30\right)\)

\(=-15+13-30\)

\(=\left(-15-30\right)+13\)

=-45+13

=-32

b: \(225-150:\left(30+3^2\cdot5\right)\)

\(=225-150:\left(30+9\cdot5\right)\)

\(=225-150:75\)

=225-2

=223

c: \(18\cdot64+18\cdot36-1200\)

\(=18\left(64+36\right)-1200\)

\(=18\cdot100-12\cdot100\)

\(=6\cdot100=600\)

d: \(80-\left[130-\left(12-4\right)^2\right]\)

\(=80-130+\left(12-4\right)^2\)

\(=-50+8^2=64-50=14\)

a) 4x^2 ( 5x^3 - 2x + 3 )

= 20x^5 - 8x^3 + 12x^2

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

= 6x^2 - 4x + 15x - 10

= 6x^2 + 11x - 10

c) ( 5x + y )^2

= ( 5x )^2 + 2 . 5x . y + y^2

= 24x^2 + 10xy + y^2