3:

a:Các tia trên hình là Ax,Ay,Bx,By,Cx,Cy

=>Có 6 tia

b: AB<AC

=>B nằm giữa A và C

=>AB+BC=AC

=>BC=4cm

c: AI=3/2=1,5cm

CI=7-1,5=5,5cm

\(a,a^2-10a+25=\left(a-5\right)^2\\ b,4x^2+4x+1=\left(2x+1\right)^2\\ c,4x^2-9=\left(2x-3\right)\left(2x+3\right)\\ d,x^3+3x^2+3x+1=\left(x+1\right)^3\\ e,a^3-3a^2b+3ab^2-b^3=\left(a-b\right)^3\\ f,y^3+8=\left(y+2\right)\left(y^2-2y+4\right)\\ g,27x^3-1=\left(3x-1\right)\left(9x^2+3x+1\right)\)

\(a^2-10x+25=\left(a-5\right)^2\)

b/ \(4x^2+4x+1=\left(2x+1\right)^2\)

c/ \(4b^2-9=\left(2b-3\right)\left(2b+3\right)\)

d/ \(x^3+3x^2+3x+1=\left(x+1\right)^3\)

e/ \(a^3-3a^2b+3ab^2-b^3=\left(a-b\right)^3\)

f/ \(y^3+8=\left(y+2\right)\left(y^2-2y+4\right)\)

g/ \(27x^3-1=\left(3x-1\right)\left(9x^2+3x+1\right)\)

Đề bài thiếu thì phải

32, A

33,B

34,C

\(n_{HCl}=0,1.0,2=0,02\left(mol\right)\\ n_{H_2SO_4}=0,1.0,1=0,01\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{NaOH}=n_{HCl}+2.n_{H_2SO_4}=0,04\left(mol\right)\\ V_{\text{dd}NaOH}=\dfrac{0,04}{0,5}=0,08\left(l\right)\\ \Rightarrow Ch\text{ọn}.A\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

P₂O₅ + 3H₂O --> 2H₃PO₄

=> m_H3PO4 = 0,1.98 = 9,8(g)

\(\overrightarrow{AB}-\overrightarrow{DC}+\overrightarrow{BC}-\overrightarrow{AD}=\left(\overrightarrow{AB}+\overrightarrow{BC}\right)-\left(\overrightarrow{AD}+\overrightarrow{DC}\right)\)

\(=\overrightarrow{AC}-\overrightarrow{AC}=\overrightarrow{0}\)