kiểm tra lại câu hỏi nhé bạn.

\(-11+23\cdot\left(-21\right)\)

\(=-11-483\)

\(=-\left(11+483\right)\)

\(=-494\)

\(20-\left[30-\left(5-1\right)^2\right]\)

\(=20-\left(30-4^2\right)\)

\(=20-\left(30-16\right)\)

\(=20-14\)

\(=6\)

\(2^3\cdot17-2^3\cdot14\)

\(=2^3\cdot\left(17-14\right)\)

\(=2^3\cdot3\)

\(=8\cdot3\)

\(=24\)

mik cảm ơn

a) \(x+5=20-\left(12-7\right)\)

\(\Rightarrow x+5=20-5\)

\(\Rightarrow x+5=15\)

\(\Rightarrow x=15-5\)

\(\Rightarrow x=10\)

b) \(15-\left(3+2x\right)=2^2\)

\(\Rightarrow3+2x=15-4\)

\(\Rightarrow3+2x=11\)

\(\Rightarrow2x=11-3\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=\dfrac{8}{2}\)

\(\Rightarrow x=4\)

c) \(-11-\left(19-x\right)=50\)

\(\Rightarrow19-x=-11-50\)

\(\Rightarrow19-x=-61\)

\(\Rightarrow x=61+19\)

\(\Rightarrow x=80\)

d) \(159-\left(25-x\right)=43\)

\(\Rightarrow25-x=159-43\)

\(\Rightarrow25-x=116\)

\(\Rightarrow x=25-116\)

\(\Rightarrow x=-91\)

e) \(\left(79-x\right)-43=-\left(17-52\right)\)

\(\Rightarrow\left(79-x\right)-43=52-17\)

\(\Rightarrow79-x-43=35\)

\(\Rightarrow36-x=35\)

\(\Rightarrow x=1\)

f) \(\left(7+x\right)-\left(21-13\right)=32\)

\(\Rightarrow7+x-8=32\)

\(\Rightarrow x-1=32\)

\(\Rightarrow x=32+1\)

\(\Rightarrow x=33\)

g) \(-x+20=-15+8+13\)

\(\Rightarrow-x+20=6\)

\(\Rightarrow x=20-6\)

\(\Rightarrow x=14\)

h) \(-\left(-x+13-142\right)+18=55\)

\(\Rightarrow x-13+142+18=55\)

\(\Rightarrow x+147=55\)

\(\Rightarrow x=55-147\)

\(\Rightarrow x=-92\)

\(17\cdot85+15\cdot17-120\)

\(=17\cdot\left(85+15\right)-120\)

\(=17\cdot100-120\)

\(=1700-120\)

\(=1580\)

mình cảm ơn

:)))

\(\left(2+4+6+...+2n\right)-53=103\)

\(\Rightarrow\left[\left(2n-2\right):2+1\right]\cdot\left(2n+2\right):2-53=103\)

\(\Rightarrow\left[2\left(n-1\right):2+1\right]\cdot\left(2n+2\right):2=103+53\)

\(\Rightarrow\left(n-1+1\right)\cdot2\cdot\left(n+1\right):2=156\)

\(\Rightarrow n\cdot\left(n+1\right)=156\)

\(\Rightarrow n\cdot\left(n+1\right)=12\cdot13\)

\(\Rightarrow n=12\)

Vậy: n = 12

\(A=1+3+3^2+...+3^{2021}\\3\cdot A=3\cdot(1+3+3^2+...+3^{2021})\\3\cdot A=3+3^2+3^3+...+3^{2022}\\3A-A=(3+3^2+3^3+...+3^{2022})-(1+3+3^2+...+3^{2021})\\2A=3+3^2+3^3+...+3^{2022}-1-3-3^2-...-3^{2021}\\2A=3^{2022}-1\\\Rightarrow A=\dfrac{3^{2022}-1}{2}\)

`#3107.101107`

\(A=1+3+3^2+...+3^{2021}\)

\(3A=3+3^2+3^3+...+3^{2021}\)

\(3A-A=\left(3+3^2+3^3+...+3^{2021}\right)-\left(1+3+3^2+...+3^{2021}\right)\)

\(2A=3+3^2+3^3+...+3^{2021}-1-3-3^2-...-3^{2021}\)

\(2A=3^{2021}-1\)

\(A=\dfrac{3^{2021}-1}{2}\)

Vậy, \(A=\dfrac{3^{2021}-1}{2}.\)

x + 5 chia hết cho x + 3

⇒ x + 3 + 2 chia hết cho x + 3

⇒ x + 3 chia hết cho x + 3 và 2 chia hết cho x + 3

⇒ x + 3 ∈ Ư(2) 

Mà: Ư(2) = {1; -1; 2; -2}

⇒ x + 3 ∈ {1; -1; 2; -2}

⇒ x ∈ {-2; -4; -1; -5}