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

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

a) -0,257 + 0,019 

= -0,238 ≃ -0,24 

b) 2,13 - 2,16 x 0,2 

= 2,13 - 0,432 

= 1,698 ≃ 1,7 

c) 1,213 + 1,18 : (-0,2)

= 1,213 + 1,18 x (-5) 

= 1,213 - 5,9

= -4,687 ≃ -4,69

\(A=2^2+2^4+2^6+...+2^{200}\\ 2^2A=2^4+2^6+...+2^{202}\\ 4A-A=\left(2^4+2^6+2^8+...+2^{202}\right)-\left(2^2+2^4+2^6+...+2^{200}\right)\\ 3A=2^{202}-2^2\) 

\(=>3A+4=2^{202}-2^2+4=2^{202}-4+4=2^{202}\) 

\(=>2^{202}=4^n\\ =>2^{202}=\left(2^2\right)^n\\ =>2^{202}=2^{2n}\\ =>2n=202\\ =>n=101\)

\(42\cdot136+272\cdot15+28\cdot68\cdot2\)

\(=42\cdot136+30\cdot136+28\cdot136\)

\(=136\left(42+30+28\right)\)

\(=136\cdot100=13600\)

42 x 136 + 272 x 15 + 28 x 68 x 2

= 21 x (2 x 136) + 272 x 15 + (68 x 4) x 7 x 2 

= 21 x 272 + 272 x 15 + 272 x 14

= 272 x (21 + 15 + 14)

= 272 x 50

= 13600

Tỉ số giữa Số học sinh khá và cả lớp là:

\(\dfrac{1}{2}\left(1-\dfrac{1}{4}\right)=\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{3}{8}\)

Số học sinh trung bình chiếm: \(1-\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{5}{8}-\dfrac{2}{8}=\dfrac{3}{8}\)(số học sinh cả lớp)

Số học sinh cả lớp là \(24:\dfrac{3}{8}=24\cdot\dfrac{8}{3}=64\left(bạn\right)\)

\(4-\dfrac{2+\dfrac{1}{2}}{2-\dfrac{1}{2}}\\ =4-\dfrac{\dfrac{4}{2}+\dfrac{1}{2}}{\dfrac{4}{2}-\dfrac{1}{2}}\\ =4-\dfrac{5}{2}:\dfrac{3}{2}\\ =4-\dfrac{5}{3}\\ =\dfrac{12}{3}-\dfrac{5}{3}\\ =\dfrac{7}{3}\)

\(4-\dfrac{2+\dfrac{1}{2}}{2-\dfrac{1}{2}}=4-\dfrac{\dfrac{5}{2}}{\dfrac{3}{2}}\)

\(=4-\dfrac{5}{2}:\dfrac{3}{2}=4-\dfrac{5}{2}\cdot\dfrac{2}{3}\)

\(=4-\dfrac{5}{3}=\dfrac{12}{3}-\dfrac{5}{3}=\dfrac{7}{3}\)