\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

= \(\dfrac{337.2023}{2}\)

= \(\dfrac{\text{681751}}{2}\)

a) P(x)=4x²-6x+a; Q(x)=x-3

Lấy P(x):Q(x)=4x-6 dư a+30

Vậy để P(x)⋮Q(x) ⇒ a+30=0 ⇒ a=-30

b) P(x)=2x²+x+a; Q(x)=x+3

Lấy P(x):Q(x)=2x-7 dư a+21

Vậy để P(x)⋮Q(x) ⇒ a+21=0 ⇒ a=-21

c) P(x)=x³+ax²-4; Q(x)=x²+4x+4

Lấy P(x):Q(x)=x+a-4 dư -4(a-5)x+12

Vậy để P(x)⋮Q(x) ⇒ -4(a-5)x+12=0 ⇒ (a-5)x=3

⇒ a-5 ϵ {-1;1;-3;3} (a ϵ Z)

⇒ a ϵ {4;6;2;8}

d) P(x)=2x²+ax+1; Q(x)=x-3

Lấy P(x):Q(x)=2x+a+6 dư 3a+19

Vậy để P(x)⋮Q(x) ⇒ 3a+19=0 ⇒ a=-19/3

e) P(x)=ax⁵+5x⁴-9; Q(x)=x-1

Lấy P(x):Q(x)=ax⁴+(a-5)x³+(a-5)x²+(a-5)x+1 dư a-4

Vậy để P(x)⋮Q(x) ⇒ a-4=0 ⇒ a=4

f) P(x)=6x³-x²-23x+a; Q(x)=2x+3

Lấy P(x):Q(x)=3x²-5x-4 dư a+12

Vậy để P(x)⋮Q(x) ⇒ a+12=0 ⇒ a=-12

g) P(x)=x³-6x²+ax-6 Q(x)=x-2

Lấy P(x):Q(x)=x²-2x+a-4 dư 2(a-4)-6

Vậy để P(x)⋮Q(x) ⇒ 2(a-4)-6=0 ⇒ a=7

Bài h có a,b bạn xem lại đề

\(1-\dfrac{3}{2}+2-\dfrac{5}{2}+...+2003-\dfrac{4007}{2}+2004\)

= \(\left(-\dfrac{1}{2}\right)+\left(-\dfrac{1}{2}\right)+...\left(-\dfrac{1}{2}\right)+2004\)

= \(2003.\left(-\dfrac{1}{2}\right)+2004\)

= \(-\dfrac{4006}{2}+\dfrac{4008}{2}\)

= \(\dfrac{2}{2}\)

= \(1\)

Vì 4<n≤7 và nϵN

⇒ n ϵ {5;6;7}

⇒ K=(n-1).(2n-1) ϵ {36;55;78}

\(f\left(x\right)=0\Leftrightarrow x^2+3=0\)

⇔ Vô nghiệm để đa thức f(x)=0 (vì x²≥0⇒x²+3>0)

\(-\dfrac{16}{18}.\dfrac{45}{32}-2,5=-\dfrac{5}{2}-\dfrac{5}{2}=-\dfrac{10}{2}=-5\)

Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :

(a+b):2=36⇒a+b=72

mà b=17

Nên a=72-17=55

Bài 2 :

a) 4567+y:34=10987

⇒ y:34=10987-4567

⇒ y:34=6420

⇒ y=6420x34

⇒ y=218280

b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)

\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)

\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)

\(\Rightarrow y=\dfrac{3}{4}\)

Bài 3 :

\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)

= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)

\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)

\(\left(1+\dfrac{1}{2}\right):\left(1+\dfrac{1}{3}\right):\left(1+\dfrac{1}{4}\right)\)

\(\left(\dfrac{2}{2}+\dfrac{1}{2}\right):\left(\dfrac{3}{3}+\dfrac{1}{3}\right):\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\)

\(\dfrac{3}{2}x\dfrac{3}{4}:\dfrac{5}{4}\)

\(\dfrac{9}{8}:\dfrac{5}{4}\)

\(\dfrac{9}{8}x\dfrac{4}{5}\)

\(\dfrac{9}{10}\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\Rightarrow100x+\left(1+2+...100\right)=5750\)

\(\Rightarrow100x+\dfrac{100.\left(100+1\right)}{2}=5750\)

\(\Rightarrow100x+50.101=5750\)

\(\Rightarrow100x+5050=5750\)

\(\Rightarrow100x=5750-5050\)

\(\Rightarrow100x=700\)

\(\Rightarrow x=7\)

a) 

Ta có:$(-123123)/(164164)=(-123\ast 1001)/(164\ast 1001)=(-628)/942$

Dạng chung của các số hữu tỉ bằng (-628628)/(942942)(−123123)/(164164) là $(-123)/(164)\ast a$(a là số bất kì).

b) 

Ta có:$(-434343)/(868686)=(-434\ast 1001)/(868\ast 1001)=(-434)/868$

Dạng chung của các số hữu tỉ bằng (-628628)/(942942)(−434343)/(868686) là (-628)/(942)*a(−434)/(868)$\ast a$(a là số bất kì).