cho tớ hỏi giữa 9 và 11 là kí hiệu j vậy ak

lưu ý:9 11 là 9/11 nhé

\(B=\left[\left(-0,5\right):0,07-0,2:0,07\right]\cdot1,5-2024^0\\ =\left[\left(-0,5\right):\dfrac{7}{100}-0,2:\dfrac{7}{100}\right]\cdot1,5-1\\ =\left[\left(-0,5\right)\cdot\dfrac{100}{7}-0,2\cdot\dfrac{100}{7}\right]\cdot1,5-1\\ =\dfrac{100}{7}\cdot\left(-0,5-0,2\right)\cdot1,5-1\\ =\dfrac{100}{7}\cdot-0,7\cdot1,5-1\\ =\dfrac{100}{7}\cdot\dfrac{-7}{10}\cdot1,5-1\\ =-10\cdot1,5-1\\ =-15-1\\ =-16\)

cứu với

a) \(\left(-32\right)^9=-32^9=-\left(2^5\right)^9=-2^{45}\)

\(\left(-16\right)^{13}=-16^{13}=-\left(2^4\right)^{13}=-2^{52}\)

Vì \(2^{45}< 2^{52}=>-2^{45}>-2^{52}\)

b) \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

Vì \(243>125=>243^{10}>125^{10}>\left(-3\right)^{50}>\left(-5\right)^{30}\) 

c) \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)

\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)

Vì: \(\dfrac{1}{8}>\dfrac{1}{9}=>\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}=>\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)

d) 

\(\left(\dfrac{1}{5}\right)^{199}>\left(\dfrac{1}{5}\right)^{200}=\left[\left(\dfrac{1}{5}\right)^2\right]^{100}\\ =\left(\dfrac{1}{25}\right)^{100}>\left(\dfrac{1}{27}\right)^{100}=\left[\left(\dfrac{1}{3}\right)^3\right]^{100}=\left(\dfrac{1}{3}\right)^{300}\) 

\(32< 2^n< 128\\ =>2^5< 2^n< 2^7\\ =>5< n< 7\)

Vì n là số nguyên nên n=6

b) \(2.16\ge2^n>4\\ =>2.2^4\ge2^n>2^2\\ =>2^5\ge2^n>2^2\\ =>5\ge n>2\)

Vì n là số nguyên nên \(n\in\left\{3;4;5\right\}\)

\(8^{12}=\left(8^3\right)^4=512^4\\ 12^8=\left(12^2\right)^4=144^4\\ \)

Nhận thấy: \(512^4>144^4\Rightarrow8^{12}>12^8\)

\(8^{12}=\left(2^3\right)^{12}=2^{36}\)

\(12^8=\left(2^2\cdot3\right)^8=\left(2^2\right)^8\cdot3^8\\ =2^{16}\cdot3^8< 2^{16}\cdot4^8=2^{16}\cdot\left(2^2\right)^8=2^{16}\cdot2^{16}=2^{32}< 2^{36}\) 

=> \(12^8< 8^{12}\)

\(1+3+5+...+x=36\\\left[ \left(x-1\right):2+1\right]\cdot\left(x+1\right):2=36\\ \dfrac{x-1+2}{2}\cdot\dfrac{x+1}{2}=36\\ \dfrac{x+1}{2}\cdot\dfrac{x+1}{2}=36\\ \dfrac{\left(x+1\right)^2}{4}=36\\ \left(x+1\right)^2=36\cdot4=144\)

TH1: x + 1 = 12 => x = 11 

TH2: x + 1 = -12 => x = -13 

Vì: x phải lớn hơn 0 => x = 11 

1+3+5+...+x =36

1+3+5+7+9+11 =36

\(1)2\cdot3\cdot4+4\cdot5\cdot6\\ =4\cdot\left(2\cdot3+5\cdot6\right)⋮4\)

\(2)3\cdot5\cdot7-1\cdot3\cdot5\\ =5\cdot\left(3\cdot7-1\cdot5\right)⋮5\) 

\(3)4\cdot6\cdot8-2\cdot3\cdot4\\ =4\cdot\left(6\cdot8-2\cdot3\right)⋮4\)

\(4)4\cdot6\cdot8+11\cdot12\cdot13\\ =\left(4\cdot6\right)\cdot8+11\cdot12\cdot13\\ =24\cdot8+11\cdot12\cdot13\\ =12\cdot2\cdot8+11\cdot12\cdot13\\ =12\cdot\left(2\cdot8+11\cdot13\right)⋮12\)

\(5)5\cdot6\cdot7+33\cdot34\cdot35\\ =\left(5\cdot7\right)\cdot6+33\cdot34\cdot35\\ =35\cdot6+33\cdot34\cdot35\\ =35\cdot\left(6+33\cdot34\right)⋮35\)

\(6)43\cdot45\cdot47-9\cdot7\cdot5\\ =43\cdot45\cdot47-\left(9\cdot5\right)\cdot7\\ =43\cdot45\cdot47-45\cdot7\\ =45\cdot\left(43\cdot47-7\right)⋮45\)

\(7)125\cdot13+50\cdot14\\ =25\cdot5\cdot13+25\cdot2\cdot14\\ =25\cdot\left(5\cdot13+2\cdot14\right)⋮25\)

\(8)125\cdot13+50\cdot14\)

Có: 125*13 ⋮ 13 

Mà: \(50\cdot14=5^2\cdot2\cdot2\cdot7=5^2\cdot2^2\cdot7\) => không chia hết cho 13

=> 125*13 + 50*14 không chia hết cho 13

\(\left(2x-4\right)\left(6-3y\right)=4\)

\(x,y\in Z=>2x-4;6-3y\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\) 

Mà: \(2x-4\) luôn chẵn 

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

Ta có bảng: 

=> Không có cặp x,y nguyên thỏa mãn 

Giải giúp với mn ơiii

ủa 16-4=12, chứ đâu phải bằng 10 đâu