\(P\left(3\right)=3^2-2\cdot3-3=0\)

=>x=3 là nghiệm

\(P\left(-1\right)=\left(-1\right)^2-2\cdot\left(-1\right)-3=0\)

=>x=-1 là nghiệm

Xét x=-1

=> P = 0 ( nhận )

Xét x=0 

=> P = -3 ( loại )

Xét x=1

=> P = -4 ( loại )

Xét x=3

=> P = 0 ( nhận )

Vậy x=-1;3 là nghiệm của đa thức

=>x=3 là nghiệm

=>x=-1 là nghiệm

bằng 11805000 m2

20500 nha bụn

3.

\(=\left[\frac{15(\sqrt{6}-1)}{(\sqrt{6}-1)(\sqrt{6}+1)}+\frac{4(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)}-\frac{12(3+\sqrt{6})}{(3-\sqrt{6})(3+\sqrt{6})}\right].\frac{1}{\sqrt{6}+11}\)

\(=\left[\frac{15(\sqrt{6}-1)}{5}+\frac{4(\sqrt{6}+2)}{2}-\frac{12(3+\sqrt{6})}{3}\right].\frac{1}{\sqrt{6}+11}\)

\(=[3(\sqrt{6}-1)+2(\sqrt{6}+2)-4(3+\sqrt{6})].\frac{1}{\sqrt{6}+11}=\frac{\sqrt{6}-11}{\sqrt{6}+11}\)

\(=\frac{(\sqrt{6}-11)^2}{(\sqrt{6}-11)(\sqrt{6}+11)}=\frac{(\sqrt{6}-11)^2}{-115}\)

1: Ta có: \(\dfrac{20\sqrt{300}+15\sqrt{675}-10\sqrt{75}}{\sqrt{15}}\)

\(=\dfrac{200\sqrt{3}+375\sqrt{3}-50\sqrt{3}}{\sqrt{15}}\)

\(=\dfrac{525\sqrt{3}}{\sqrt{15}}=105\sqrt{5}\)

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

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

=-1+5

=4

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

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\cdot\left(\sqrt{6}+11\right)\)

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

=6-121

=-115

36/B
37/C
38/D
39/D
40/C

36.B
37.C
38.D
39.D
40.C

b

c

d

d

c

Ý C

\(\sqrt{\left(\left(xN-xM\right)^2+\left(yN-yM\right)^2\right)}\)·=\(\sqrt{\left(4-1\right)^2+\left(2+2\right)^2}\)  = 5

A

D

D

mk ko thấy Anh Mai