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: \(\dfrac{2^{100}+2^{101}+2^{102}}{2^{97}+2^{98}+2^{99}}\)

\(=\dfrac{2^{100}\left(1+2+2^2\right)}{2^{97}\left(1+2+2^2\right)}\)

\(=\dfrac{2^{100}}{2^{97}}=2^3=8\)

thankss

\(\dfrac{1}{8}-\dfrac{4}{8}=\dfrac{-3}{8}\)

-3/8

(101+102+...+200)+(-1-2-3-...-100)

=(101-1)+(102-2)+...+(200-100)

=100+100+...+100

=100*100=10000

\(3\cdot3^{10}=3^{1+10}=3^{11}\)

\(3\cdot3^{10}=3^{1+10}=3^{11}\)

Theo nhân hai luỹ thừa cùng cơ số: 

Ta có: \(a^m\cdot a^n=a^{m+n}\)

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}

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: \(2x+7⋮x-3\)

=>\(2x-6+13⋮x-3\)

=>\(13⋮x-3\)

=>\(x-3\in\left\{1;-1;13;-13\right\}\)

=>\(x\in\left\{4;2;16;-10\right\}\)

Đề đâu em?

giúp gì bn

`@` `\text {Ans}`

`\downarrow`

`14*43*86*43`

`= 14*43*43*2*43`

`= 43^3*14*2`

`= 43^3*28`

`= 2226196`