Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

\(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-\)\(\sqrt{75}\)

\(=\frac{1}{2}.6\sqrt{2}+\frac{3}{4}.4\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)

\(=3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)

\(=12\sqrt{2}-2\sqrt{3}\)

Trước \(\sqrt{75}\)là dấu " - " hay dấu " --  " vậy? Nếu là dấu " -- " thì \(--\sqrt{75}\Rightarrow+\sqrt{75}\)nha

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

a)\(\left(-3\right).8.\left(-2\right).5=\left(-3\right).\left(-2\right).8.5\)

                                          \(=6.40=240\)

b)\(127-18.\left(5+6\right)=127-18.11\)

                                             \(=127-198=-71\)

c)\(125-\left(-75\right)+32-\left(48+32\right)=125+75+32-48-32\)

                                                                              \(\left(125+75\right)+\left(32-32\right)-48=200-48=152\)

d)\(3.\left(-4\right).2+2.\left(-5\right)-20=-12.2+2.\left(-5\right)-20\)

                                                             \(=\left(-12-5\right).2-20=-17.2-20=-140-20=-160\)

Chúc bạn học tốt

125 - ( - 75) + 32 - ( 48 + 32)

= 125 - ( - 75) + 32 - 80

= 200 + 32 - 80

= 232 - 80

= 152

3 . ( -4)² + 2 . ( -5) - 20

= 3 . 16 + ( -10 ) - 20

= 48 + ( -10 ) - 20

= 38 - 20

= 18

a)=125+75+32-48-32

=(125+75)+(32-32)-48

=200+0-48

=152

b)=3.16-10-20

=48-(10+20)

=48-30

=18

125 - (-75) + 32 - (48 + 32)

=125 + 75 + 32 - 48 - 32

=200 + 32 - 32 - 48

=200 - 48 

=152+

3 . (-4)^2 + 2. (-5) - 20

=3 . 16 + 2. (-5) - 20

=48 + (-10) - 20

=18 

a) \(P=\dfrac{\sqrt{3}+\sqrt{6}}{1+\sqrt{2}}=\dfrac{\left(\sqrt{3}+\sqrt{6}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)

                             \(=\dfrac{\sqrt{3}-\sqrt{6}+\sqrt{6}-\sqrt{12}}{1-2}=\sqrt{12}-\sqrt{3}\)

b) \(Q=\left(\sqrt{75}-\dfrac{3}{2}:\sqrt{3}-\sqrt{48}\right)\cdot\sqrt{\dfrac{16}{3}}\)

        \(=\left(5\sqrt{3}-\dfrac{3}{2}\cdot\dfrac{1}{\sqrt{3}}-4\sqrt{3}\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\sqrt{3}\left(5-\dfrac{1}{2}-4\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\left(1-\dfrac{1}{2}\right)\cdot4=2\)

a) \(125-\left(-75\right)+32-\left(48+32\right)\)

\(=125+75+32-48-32\)

​\(=200+\left(32-32\right)-48\)​

\(=200+0-48\)

\(=200-48\)

\(=152\)

b) \(3.\left(-4\right)^2+2.\left(-5\right)-20\)

\(=3.16-10-20\)

\(=48-30\)

\(=18\)

c) \(-1685+\left(-2365\right)+569-1685\)

\(=\left(-1685-1685\right)-2365+569\)

\(=-3370-2365+569\)

\(=-5735+569\)

\(=-5166\)

Câu 12: B

Câu 13: C

Câu 14: A

C12 B

C13 C

C14 A

12.B     13. C         14.  A

(-5).8.(-2).(-125)

=[(-5).(-2)].[8.(-125)]

=10.(-1000)

=-10000