a; \(2x^3\left(5x-1\right)=2x^3\cdot5x-2x^3\cdot1=10x^4-2x^3\)

b: (x-2)(2x-5)

\(=2x^2-5x-4x+10\)

\(=2x^2-9x+10\)

c: \(-2x^2\left(3x-1\right)=-2x^2\cdot3x+2x^2\cdot1=-6x^3+2x^2\)

d: (x-y)(x+y)\(=x^2+xy-xy-y^2=x^2-y^2\)

e: \(3x^4\cdot\left(2x-y\right)=3x^4\cdot2x-3x^4\cdot y=6x^5-3x^4y\)

f: (x-5)(x+5)=\(x^2+5x-5x-25=x^2-25\)

g: \(-2x^2\left(4x^2-2x+1\right)=-2x^2\cdot4x^2+2x^2\cdot2x-2x^2\cdot1=-8x^3+4x^3-2x^2\)

h: (a+3)(a+5)

\(=a^2+5a+3a+15=a^2+8a+15\)

Chọn C

\(2.16\ge2^n>4\)

\(2.2^4\ge2^n>2^2\)

\(2^5\ge2^n>2^2\)

=> \(n\in\left\{3,4,5\right\}\)

Vậy: \(n\in\left\{3,4,5\right\}\)

CT : CnH2n+1OH

nCO2 = 13.2/44 = 0.3 (mol) 

nH2O = 7.2/18 = 0.4 (mol) 

CnH2n+1OH + 3n/2O2  -to-> nCO2 + (n+1)H2O

...........................................n............n+1

...........................................0.3...........0.4

=> 0.4n = 0.3(n+1) 

=> n = 3 

CTPT : C3H7OH 

CTCT : 

CH3 - CH2 - CH2 - OH : propan - 1 - ol 

CH3 - CH(CH3)-OH : propan - 2 - ol 

a: Xét ΔABC có \(\widehat{B}+\widehat{C}=90^0\)

nên ΔABC vuông tại A

Xét ΔABC vuông tại A có 

\(AB=BC\cdot\sin\widehat{C}\)

\(=40\cdot\sin50^0\simeq30,64\left(cm\right)\)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(AB^2+AC^2=BC^2\)

\(\Leftrightarrow AC^2=40^2-30.64^2=661.1904\)

hay \(AC\simeq25,71cm\)

g: \(x\left(x-5\right)-3\left(x-5\right)=\left(x-5\right)\left(x-3\right)\)

h: \(x\left(x-y\right)-2\left(y-x\right)=\left(x-y\right)\left(x+2\right)\)

i: \(x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)\)

k: \(m\left(x-3\right)-n\left(x-3\right)=\left(x-3\right)\left(m-n\right)\)

l: \(5x-10=5\left(x-2\right)\)

\(a)5m-5n=5(m-n)\\b) -2x-2y=-2(x+y)\\c)-7+7y=-7(1-y)\\d)10x^3-15x^2=5x^2(2x-3)\\e) x^2-xy=x(x-y)\\f)9x^4-6x^2=3x^2(3x^2-2)\\g)x(x-5)-3(x-5)=(x-3)(x-5)\\h)x(x-y)-2(y-x)=x(x-y)+2(x-y)=(x+2)(x-y)\\i)x(x+3)+5(3+x)=(x+5)(x+3)\\k)m(x-3)+n(3-x)=m(x-3)-n(x-3)=(m-n)(x-3)\\l)5x-10=5(x-2) \)