f: =>(x-3)(x-1)>=0

=>x>=3 hoặc x<=1

g: =>(x-2)(x^2+3)<0

=>x-2<0

=>x<2

h: =>x+2>=0

=>x>=-2

i: =>(x-1-x+3)/(x-3)>0

=>x-3>0

=>x>3

a)\(\dfrac{-2}{17}+\dfrac{8}{17}=\dfrac{6}{17}\)

b) \(\dfrac{-9}{16}+\dfrac{7}{24}=\dfrac{-13}{48}\)

c)\(\dfrac{1}{8}+\dfrac{-5}{6}+\dfrac{7}{-36}=\dfrac{-17}{24}+\dfrac{7}{-36}=\dfrac{-65}{72}\)

a) =\(\dfrac{-2+8}{17}=\dfrac{6}{17}\)
b) =\(\dfrac{-27}{48}+\dfrac{14}{48}=\dfrac{-27+14}{48}=\dfrac{-13}{48}\)
c) \(=\dfrac{9}{72}+\dfrac{-60}{72}+\dfrac{-14}{72}=\dfrac{9+\left(-60\right)+\left(-14\right)}{72}=\dfrac{-65}{72}\)

5 B

6 B

7 D

8 B

9 C

10 A

6. B

7. A

8. B

9. A

10.A/C ( mk ko rõ )

=3/7+2/5=15/35+14/35=29/35

=3/7+2/5=15/35+14/35=29/35

likes

Likes

likes

ko bt lm :<<

\(\dfrac{x}{z}=\dfrac{z}{y}\Rightarrow xy=z^2\)

\(VT=\dfrac{x^2+z^2}{y^2+z^2}=\dfrac{x^2+xy}{y^2+xy}=\dfrac{x\left(x+y\right)}{y\left(x+y\right)}=\dfrac{x}{y}=VP\)

Lời giải:

Ta có:

\(P=1+x+x^2+x^3+...+x^9+x^{10}\)

\(\Rightarrow xP=x+x^2+x^3+...+x^{10}+x^{11}\)

Trừ theo vế:
\(xP-P=(x+x^2+x^3+...+x^{10}+x^{11})-(1+x+x^2+...+x^{10})\)

\(\Rightarrow \)\(xP-P=x^{11}-1\) (đpcm)

P.s: Bạn lưu ý lần sau nhớ viết công thức rõ ràng.