a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

cứu 

okee laaa

quá chuẩn luôn !!!!!!!!

NHỚ L.I.K.E cho mk nha

 a) (x+2)(x^2-2x+4)-x(x^2+2)=15 
<=> x^3 + 8 - x^3 - 2x = 15 
<=> -2x = 7 
<=> x = -7/2 

b) (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28 
<=> x^3 + 9x² + 27x + 27 - x(9x² + 6x + 1) + 8x^3 + 1 = 28 
<=> x^3 + 9x² + 27x + 27 - 9x^3 - 6x² - x + 8x^3 + 1 - 28 = 0 
<=> 3x² + 26x = 0 
<=> x(3x + 26) = 0 
Vậy x = 0 và x = -26/3 

c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0 
<=> (x² - 1)[(x² -1)² - x^4 - x² - 1] = 0 
<=> (x-1)(x+1)(x^4 - 2x² + 1 - x^4 - x² - 1 ) = 0 
<=> -(x-1)(x+1)3x² = 0 
Vậy nghiệm là x = 1 ; -1 ; 0

Biết rồi đăng lên làm j

đề bài giống này đúng ko bạn nếu đúng thì làm theo nha nhớ k cho mình

E= \(1^3+2^3+3^3+...+99^3+100^3\)

=(1-1)1(1+1)+1+(2-1)2(2-1)+2+...+)(99-1)99(99+1)+99+(100-1) 100(100+1)+100

= 1+2+1.2.3+3+2.3.4+...+100+99.100+101

= (1+2+3+..+100) +(1.2.3+2.3.4+...+99.100.101)

= 5050+25497450

=25502500

\(B=\left(8-2,25+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+1\dfrac{1}{4}\right)-\left(3+0,5-1\dfrac{2}{7}\right)\\ B=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{1}{2}+\dfrac{9}{7}\\ B=\left(8+6-3\right)+\left(-\dfrac{9}{4}-\dfrac{5}{4}\right)+\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\dfrac{1}{2}\)

\(B=11-\dfrac{7}{2}+2-\dfrac{1}{2}\\ B=\left(11+2\right)+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\\ B=13-4\\ B=9.\)

\(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)\)

\(A=\dfrac{2}{3}+\dfrac{-1}{3}\)

\(A=\dfrac{1}{3}\)

_____

( *Xem lại câu B nha bạn )

_____

\(C=\left(\dfrac{3}{4}-0,2\right).\left(0,4-\dfrac{4}{5}\right)\)

\(C=\left(\dfrac{3}{4}-\dfrac{1}{5}\right).\left(\dfrac{2}{5}-\dfrac{4}{5}\right)\)

\(C=\dfrac{11}{20}.\dfrac{-2}{5}\)

\(C=\dfrac{-11}{50}\)

\(x^3:\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)

\(\Rightarrow x^3:\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)

\(\Rightarrow x^3=\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

\(\Rightarrow x^3=\left(\dfrac{1}{2}\right)^3\)

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

\(x^3:\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\Rightarrow x^3=\dfrac{1}{2}.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}.\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^3\)

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

x3:(−21​)2=21​

⇒�3:(12)2=12⇒x3:(21​)2=21​

⇒�3=(12)2⋅12⇒x3=(21​)2⋅21​

⇒�3=(12)3⇒x3=(21​)3

⇒�=12⇒x=21​
 

Úi khó thế em mới lớp 3

Ta có: 

\(\frac{bc}{a^2+1}\le\frac{1}{4}.\frac{\left(b+c\right)^2}{a^2+b^2+a^2+c^2}\)

\(\le\frac{1}{4}.\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+b^2}\right)\)(1)

Tương tự ta có:

\(\hept{\begin{cases}\frac{ac}{b^2+1}\le\frac{1}{4}.\left(\frac{a^2}{b^2+a^2}+\frac{c^2}{b^2+c^2}\right)\\\frac{ab}{c^2+1}\le\frac{1}{4}.\left(\frac{a^2}{c^2+a^2}+\frac{b^2}{c^2+b^2}\right)\end{cases}}\)

Cộng mấy cái trên vế theo vế ta được

\(\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\le\frac{1}{4}.\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}+\frac{a^2}{b^2+a^2}+\frac{c^2}{b^2+c^2}+\frac{a^2}{c^2+a^2}+\frac{b^2}{c^2+b^2}\right)\)

\(=\frac{3}{4}\)

\(\frac{bc}{a^2+1}=\frac{bc}{a^2+b^2+a^2+c^2}\le\frac{1}{4}\left(\frac{bc}{a^2+b^2}+\frac{bc}{a^2+c^2}\right)\le\frac{1}{4}\left(\frac{bc}{2ab}+\frac{bc}{2ac}\right)\)