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\(\left(3x-4\right)^3=5^2+4.5^2\)
\(\Leftrightarrow\left(3x-4\right)^3=5^2\left(1+4\right)\)
\(\Leftrightarrow\left(3x-4\right)^3=5^3\)
\(\Leftrightarrow3x-4=5\Leftrightarrow3x=9\Leftrightarrow x=3\)
Ta có: \(\left(3x-4\right)^3=5^2+4\cdot5^2\)
\(\Leftrightarrow3x-4=5\)
hay x=3
Lời giải:
$A=7+(7^2+7^3+7^4+7^5)+(7^6+7^6+7^8+7^9)+....+(7^{2018}+7^{2019}+7^{2020}+7^{2021})$
$=7+7^2(1+7+7^2+7^3)+7^6(1+7+7^2+7^3)+....+7^{2018}(1+7+7^2+7^3)$
$=7+(1+7+7^2+7^3)(7^2+7^6+....+7^{2018}$
$=7+400(7^2+7^6+....+7^{2018})$
Dễ thấy $400(7^2+7^6+....+7^{2018})$ tận cùng là $0$
Do đó $A$ tận cùng là $7$
\(3\left(x+2\right)^3-1^{2019}=5\cdot4^2\)
\(\Leftrightarrow3\left(x+2\right)^3=5\cdot16+1=81\)
\(\Leftrightarrow x+2=3\)
hay x=1
\(4^{15}.9^{15}< 2^n.3^n< 18^{16}.2^{16}\)
⇒\(\left(4.9\right)^{15}< \left(2.3\right)^n< \left(18.2\right)^{16}\)
⇒\(\left(6^2\right)^{15}< 6^n< \left(6^2\right)^{16}\)
⇒\(6^{30}< 6^n< 6^{32}\)
⇒\(6^n=6^{31}\)
⇒n=31
\(4^{15}\cdot9^{15}< 2^n\cdot3^n< 18^{16}\cdot2^{16}\\ \Leftrightarrow\left(4\cdot9\right)^{15}< \left(2\cdot3\right)^n< \left(18\cdot2\right)^{16}\\ \Leftrightarrow36^{15}< 6^n< 36^{16}\\ \Leftrightarrow6^{30}< 6^n< 6^{32}\\ \Leftrightarrow n=31\)
BN THAM KHẢO:
Lời giải 1:10−2n⋮n−210−2n⋮n−2
⇔2n−10⋮n−2⇔2n−10⋮n−2
⇔2(n−2)−6⋮n−2⇔2(n−2)−6⋮n−2
⇔6⋮n−2⇔6⋮n−2
Ta có bảng
n - 2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
n | -4 | -1 | 0 | 1 | 3 | 4 | 5 | 8 |
Vậy n∈{−4;−1;0;1;3;4;5;8}
Lời giải 2:
Ta có :
10-2n = -2n+10 = -2n+4 + 6 = -2.(n-2) + 6
Vì -2.(n-2) chia hết cho n-2
=> để 10-2n chia hết cho n-2
=> 6 chia hết cho n - 2
=> n-2 ∈ Ư(6) = {-1;1;2;-2;3;-3;6;-6}
=> n ∈ {1;3;4;0;5;-1;8;-4}
BN CHỌN CÁCH NÀO CŨNG ĐC!
\(\left(10-2n\right)⋮\left(n-2\right)\)
\(\Rightarrow-2\left(n-2\right)+6⋮\left(n-2\right)\)
\(\Rightarrow\left(n-2\right)\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
\(\Rightarrow n\in\left\{3;1;4;0;5;1;8;-4\right\}\)
c)\(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)....\left(1+\dfrac{1}{2020}\right)\left(1+\dfrac{1}{2021}\right)\)
\(=\left(\dfrac{1.2}{1.2}+\dfrac{1}{2}\right)\left(\dfrac{1.3}{1.3}+\dfrac{1}{3}\right)...\left(\dfrac{1.2021}{1.2021}+\dfrac{1}{2021}\right)\)
\(=\dfrac{3}{1.2}\cdot\dfrac{4}{1.3}\cdot\cdot\cdot\cdot\dfrac{2022}{1.2021}\)
\(=\dfrac{3.4.5...2022}{\left(1.1.1....1\right)\left(2.3.4...2021\right)}\)
\(=\)\(\dfrac{3.4.5...2022}{2.3.4...2021}\)
\(=\dfrac{2022}{2}=1011\)
\(d\))\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{199}\right)\left(1-\dfrac{1}{200}\right)\)
\(=\left(\dfrac{2}{1.2}-\dfrac{1}{1.2}\right)\left(\dfrac{3}{1.3}-\dfrac{1}{1.3}\right)....\left(\dfrac{200}{1.200}-\dfrac{1}{1.200}\right)\)
\(=\dfrac{1.2.3....199}{\left(1.1.1....1\right).\left(2.3.4....200\right)}\)
\(=\dfrac{1.2.3...199}{2.3.4...200}\)
Nếu mik làm sai mong bạn thông cảm
\(=\left(\dfrac{1}{49}-\dfrac{1}{9}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)=0\)
\(\dfrac{-13}{8}+\dfrac{-5}{9}+\dfrac{26}{26}-\dfrac{13}{9}\)
= \(\left(\dfrac{-13}{8}+\dfrac{26}{16}\right)+\left(\dfrac{-5}{9}-\dfrac{13}{9}\right)\)
= \(\left(\dfrac{-26}{16}+\dfrac{26}{26}\right)+\left(\dfrac{-18}{9}\right)\)
= \(0+\left(-2\right)\)
= \(-2\)
\(\left(\dfrac{-13}{8}-\dfrac{26}{16}\right)+\left(\dfrac{-5}{9}-\dfrac{13}{9}\right)=\left(\dfrac{-13}{8}-\dfrac{13}{8}\right)+\dfrac{-18}{9}=0+\left(-2\right)=-2\)
Bài 3:
$\frac{5}{6}, \frac{3}{4}, \frac{-1}{12}, \frac{-2}{9}, \frac{-5}{8}$
Bài 4:
$\frac{8}{18}=\frac{4}{9}$
$\frac{14}{21}=\frac{2}{3}=\frac{6}{9}$
$\frac{10}{25}=\frac{2}{5}$
$\frac{44}{77}=\frac{4}{7}$
$\frac{5}{10}=\frac{1}{2}$
Sắp xếp theo thứ tự tăng dần là: $\frac{10}{25}, \frac{8}{18}, \frac{5}{10}, \frac{44}{77}, \frac{14}{21}$