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10 tháng 1
Bài 1:
\(A=3+3^2+...+3^{100}\)
=>\(3\cdot A=3^2+3^3+...+3^{101}\)
=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2A=3^{101}-3\)
=>\(2A+3=3^{101}\)
mà \(2A+3=3^n\)
nên n=101
Bài 2:
a: \(M=3+3^2+3^3+3^4+...+3^{100}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{98}\right)⋮12\)
=>\(M=4\cdot3\cdot\left(1+3^2+...+3^{98}\right)⋮4\)
b: \(M=3+3^2+...+3^{100}\)
=>\(3M=3^2+3^3+...+3^{101}\)
=>\(3M-M=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)
=>\(2M=3^{101}-3\)
=>\(2M+3=3^{101}\)
=>n=101
26 tháng 11 2023
Bài 3:
a: \(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
\(=\dfrac{11\cdot3^{29}-3^{30}}{3^{28}\cdot2^2}=\dfrac{3^{29}\left(11-3\right)}{3^{28}\cdot4}\)
\(=3\cdot\dfrac{8}{4}=3\cdot2=6\)
b: \(\dfrac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}\)
\(=\dfrac{2^{10}\cdot3^9\cdot3-2^{10}\cdot3^9\cdot1}{2^9\cdot3^{10}}\)
\(=\dfrac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}\)
\(=\dfrac{2}{3}\cdot2=\dfrac{4}{3}\)
c: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
Bài 4:
a: \(\dfrac{10^2+11^2+12^2}{13^2+14^2}\)
\(=\dfrac{100+121+144}{169+195}\)
\(=\dfrac{365}{365}=1\)
b: \(\dfrac{2^3\cdot9^4+9^3\cdot45}{9^2\cdot10-9^2}\)
\(=\dfrac{2^3\cdot9^4+9^3\cdot9\cdot5}{9^2\left(10-1\right)}\)
\(=\dfrac{9^4\left(2^3+5\right)}{9^3}=9\cdot\left(8+5\right)=9\cdot13=117\)
c: \(\left[\dfrac{3^{14}\cdot69+3^{14}\cdot12}{3^{16}}-7\right]:2^4\)
\(=\left(\dfrac{3^{14}\left(69+12\right)}{3^{14}\cdot9}-7\right):16\)
\(=\dfrac{\left(\dfrac{81}{9}-7\right)}{16}=\dfrac{2}{16}=\dfrac{1}{8}\)
d: \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(\dfrac{24}{3}\right)^4-\left(\dfrac{32}{16}\right)^{12}\)
\(=8^4-2^{12}\)
\(=2^{12}-2^{12}=0\)
e: \(2010^{2010}\left(7^{10}:7^8-3\cdot2^4-2^{2010}:2^{2010}\right)\)
\(=2010^{2010}\left(7^2-3\cdot16-1\right)\)
\(=2010^{2010}\cdot\left(49-48-1\right)\)
=0
f: \(\dfrac{2^{100}+2^{101}+2^{102}}{2^{97}+2^{98}+2^{99}}\)
\(=\dfrac{2^{100}\left(1+2+2^2\right)}{2^{97}\left(1+2+2^2\right)}\)
\(=\dfrac{2^{100}}{2^{97}}=2^3=8\)
17 tháng 12 2023
(101+102+...+200)+(-1-2-3-...-100)
=(101-1)+(102-2)+...+(200-100)
=100+100+...+100
=100*100=10000
TP
2
H9
HT.Phong (9A5)
CTVHS
11 tháng 1
Bài 4:
a) 2x + 7 ⋮ x + 2
⇒ 2x + 4 + 3 ⋮ x + 2
⇒ 2(x + 2) + 3 ⋮ x + 2
⇒ 3 ⋮ x + 2
⇒ x + 2 ∈ Ư(3) = {1; -1; 3; -3}
⇒ x ∈ {-1; -3; 1; -5}
b) 2x + 7 ⋮ x - 3
⇒ 2x - 6 + 13 ⋮ x - 3
⇒ 2(x - 3) + 13 ⋮ x - 3
⇒ 13 ⋮ x - 3
⇒ x - 3 ∈ Ư(13) = {1; -1; 13; -13}
⇒ x ∈ {4; 2; 16; -10}
11 tháng 1
Bài 6:
a: \(3x-13⋮x+3\)
=>\(3x+9-22⋮x+3\)
=>\(-22⋮x+3\)
=>\(x+3\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;8;-14;19;-25\right\}\)
b: \(2x+24⋮x-4\)
=>\(2x-8+32⋮x-4\)
=>\(32⋮x-4\)
=>\(x-4\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16;32;-32\right\}\)
=>\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12;36;-28\right\}\)
Bài 5:
a: \(4x+3⋮x-2\)
=>\(4x-8+11⋮x-2\)
=>\(11⋮x-2\)
=>\(x-2\in\left\{1;-1;11;-11\right\}\)
=>\(x\in\left\{3;1;13;-9\right\}\)
b: \(2x+7⋮x-3\)
=>\(2x-6+13⋮x-3\)
=>\(13⋮x-3\)
=>\(x-3\in\left\{1;-1;13;-13\right\}\)
=>\(x\in\left\{4;2;16;-10\right\}\)
Bài 4:
a: \(2x+7⋮x+2\)
=>\(2x+4+3⋮x+2\)
=>\(3⋮x+2\)
=>\(x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-1;-3;1;-5\right\}\)
b: \(2x+7⋮x-3\)
=>\(2x-6+13⋮x-3\)
=>\(13⋮x-3\)
=>\(x-3\in\left\{1;-1;13;-13\right\}\)
=>\(x\in\left\{4;2;16;-10\right\}\)
QT
1
19 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`14*43*86*43`
`= 14*43*43*2*43`
`= 43^3*14*2`
`= 43^3*28`
`= 2226196`