9999 / 36 - 15*11 = 1111 / 4 - 165 = 1111 / 4 - 660 / 4 = 451 / 4 = 112 ,75

\(A=\dfrac{cota-tana}{tana+2\cdot cota}\)

\(=\dfrac{\dfrac{cosa}{sina}-\dfrac{sina}{cosa}}{\dfrac{sina}{cosa}+2\cdot\dfrac{cosa}{sina}}\)

\(=\dfrac{cos^2a-sin^2a}{sina\cdot cosa}:\dfrac{sin^2a+2\cdot cos^2a}{sina\cdot cosa}\)

\(=\dfrac{cos^2a-sin^2a}{sin^2a+2\cdot cos^2a}\)

\(=\dfrac{1-2\cdot sin^2a}{sin^2a+2\left(1-sin^2a\right)}\)

\(=\dfrac{1-2\cdot sin^2a}{-sin^2a+2}\)

\(=\dfrac{1-2\cdot\left(\dfrac{1}{3}\right)^2}{-\left(\dfrac{1}{3}\right)^2+2}=\dfrac{1-\dfrac{2}{9}}{-\dfrac{1}{9}+2}=\dfrac{7}{9}:\dfrac{17}{9}=\dfrac{7}{17}\)

375 – 10 x 3 = 375 - 30

= 345

64 : 8 + 30 = 8 + 30

= 38

\(\hept{\begin{cases}x+y=m\\x^2+y^2=n\end{cases}\Rightarrow x^2+2xy+y^2=m^2\Rightarrow xy=\frac{m^2-n}{2}}\)

P =\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=m.\left(n-\frac{m^2-2}{2}\right)\)

\(=m.\frac{3n-m^2}{2}=\frac{3mn-m^3}{2}\)

(77+22-65)-(67+12-75)

=77+22-65-67-12+75

=(77-67)+(22-12)+(75-65)

=10+10+10

=30

=2+9+1+18+1+27+1+36+1+45+1+54+1+63+1+1

=9+9x2+9x3+9x4+9x5+9x6+9x7+9

=9x(1+2+3+4+5+6+7+1)

=9x29

Ta có \(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=2\sqrt{2\left(a^2+ab+2bc+2ca\right)}\)

\(=2\sqrt{2\left(a+b\right)\left(a+2c\right)}\le2\left(a+b\right)+\left(a+2c\right)=3a+2b+2c\)

Tương tự \(\sqrt{8b^2+56}\le2a+3b+2c;\)\(\sqrt{4c^2+7}=\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\frac{a+b+4c}{2}\)

Do vậy \(Q\ge\frac{11a+11b+12c}{3a+2b+2c+2a+3b+2c+\frac{a+b+4c}{2}}=2\)

Dấu "=" xảy ra khi và chỉ khi \(\left(a,b,c\right)=\left(1;1;\frac{3}{2}\right)\)

a) \(P=1957\)

b) \(S=19.\)

2+10+19+28+37+46+55+64

=2+9+1+18+1+17+1+36+1+45+1+54+1+63+1

=9+9x2+9x3+9x4+9x5+9x6+9x7+9

=9x(1+2+3+4+5+6+7+1)

=9x29

=2+9+1+18+1+27+1+36+1+45+1+54+1+63+1

=9+9x2+9x3+9x4+9x5+9x6+9x7+9

=9x(1+2+3+4+5+6+7+1)

=9x29

Chọn B