chắc là bé hơn

Ta có :

    \(B=4+2^2+2^3+2^4+...+2^{2016}\)

\(\Rightarrow\) \(B-4=2^2+2^3+2^4+...+2^{2016}\)

\(\Rightarrow\) \(2\left(B-4\right)=2^3+2^4+2^5+...+2^{2017}\)

\(\Rightarrow\) \(2\left(B-4\right)-\left(B-4\right)=B-4=2^{2017}-2^2\)

\(\Rightarrow\) \(B=2^{2017}-2^2+4=2^{2017}\)

\(\Rightarrow\) \(A=B=2^{2017}\)

Vậy \(A=B\)

\(M=5x^2+10y^2-2xy+4x-6y+2\)

        \(=\left(x^2-2xy+y^2\right)+\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)+1\)

      \(=\left(x-y\right)^2+\left(2x+1\right)^2+\left(3y-1\right)^2+1\ge1\)

vậy \(M\ge N\)

Tại sao có 3 con số 1 vậy pạn??? @Nguyễn_thị_huyền_anh

mình xin lỗi mình nhầm chút 

vậy M\(\ge0\)

=>  M<N

i giúp em vớiiiiii

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)

\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)

\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)

\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)

mà \(3< \dfrac{10}{3}\)

nên \(M< \dfrac{10}{3}\)

\(\left(\sqrt{7}-2\right)^2=11-4\sqrt{7}\)

\(\left(3-\sqrt{7}\right)^2=16-6\sqrt{7}=11-4\sqrt{7}+5-2\sqrt{7}\)

mà \(5-2\sqrt{7}< 0\)

nên \(\sqrt{7}-2< 3-\sqrt{7}\)

\(\dfrac{5}{-6}=\dfrac{-55}{66};\dfrac{-10}{11}=\dfrac{-60}{66}\Rightarrow\dfrac{-55}{66}>\dfrac{-60}{66}\Rightarrow\dfrac{5}{-6}>\dfrac{-10}{11}\\ \dfrac{-3}{20}=\dfrac{-45}{300};\dfrac{2}{-15}=\dfrac{-40}{300}\Rightarrow\dfrac{-45}{300}< \dfrac{-40}{300}\Rightarrow\dfrac{-3}{20}< \dfrac{2}{-15}\\ -0,305>-0,36\)

\(^\circ\) \(\dfrac{5}{-6}\) và \(\dfrac{-10}{11}\)

Ta có \(:\)

\(\dfrac{5}{-6} = \dfrac{ 5 . 11 }{ -6 . 11 } = \dfrac{ 55 }{ -66} \)

\(\dfrac{-10}{11} = \dfrac{-10 . ( -6 )}{11.(-6)} = \dfrac{60}{-66}\)

Do \(55 < 60\)

\(=> \dfrac{55}{-66} > \dfrac{60}{-66}\)

Vậy \(\dfrac{55}{-66} > \dfrac{60}{-66}\)

\(\dfrac{-3}{20}\) và \(\dfrac{2}{-15}\)

Ta có :

\(\dfrac{-3}{20} = \dfrac{-3.3}{20.3}=\dfrac{-9}{60}\)

\(\dfrac{2}{-15} = \dfrac{2.4}{-15.4}= \dfrac{8}{-60}= \dfrac{-8}{60}\)

Do \(-9 <-8\)

\(=> \dfrac{-3}{20} < \dfrac{-2}{15}\)

Vậy \( \dfrac{-3}{20} < \dfrac{-2}{15}\)

a -35/50 = -7/10

b  510/2805 = 2/11

c  119/126

B2

-2/3= -8/12 , -1/4= -3/12

-8/12<-3/12 nên -2/3<-1/4

b 2/3  5/6

12/18 và 15/18

12/18<15/18

nên 14/21<60/72

bài 1 :

a) = -7/10

b) = 510/2805 = 2/11

c) = 17/18