\(2^3+3\cdot\left(\dfrac{1}{9}\right)^0-2^{-2}\cdot4+\left[\left(-2\right)^2:\dfrac{1}{2}\right]\cdot8\)

\(=8+3\cdot1-\dfrac{1}{4}\cdot4+\left(4:\dfrac{1}{2}\right)\cdot8\)

\(=8+3-\dfrac{4}{4}+4\cdot2\cdot8\)

\(=11-1+8\cdot8\)

\(=10+64\)

\(=74\)

\(=8+3-2^{-2}\cdot2^2+\left[4\cdot2\right]\cdot8\)

=11-1+8*8

=64+10=74

=8+3−2^(−2)⋅2^2+[4⋅2]⋅8=8+3−2−2⋅22+[4⋅2]⋅8

=11-1+8*8

=64+10=74

Lời giải:

$A=9+2.3^2+2.3^3+2.3^4+...+2.3^{2023}$

$A-9=2(3^2+3^3+3^4+...+3^{2023})$

$3(A-9)=2(3^3+3^4+3^5+...+3^{2024})$

$\Rightarrow 3(A-9)-(A-9)=2(3^{2024}-3^2)$

$2(A-9)=2.3^{2024}-18$

$\Rightarrow 2A-18=2.3^{2024}-18$

$\Rightarrow A=3^{2024}\vdots 3^{2023}$ (đpcm)

Bài 1 :

Để \(x^2+5x-x^2\)

\(\Leftrightarrow5>-x^2+x\)

=>1/2x=-5/9+2/3=1/9

=>x=1/9*2=2/9

nguyễn tiến bình chép bài kìa bà con ơi

x=5

cần lời giải chi tiết ko

5 nhé

duyệt đi

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{20-18}{18.19.20}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}=\dfrac{1}{2}-\dfrac{1}{19.20}\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

bn vt đề hẳn hỏi ra dc ko bn

vt thế này mk ko hiểu nhé

^^

,,,