Chia cả tử và mẫu cho \(cosa\)

\(D=\dfrac{\dfrac{cosa}{cosa}+\dfrac{sina}{cosa}}{\dfrac{cosa}{cosa}-\dfrac{sina}{cosa}}=\dfrac{1+tana}{1-tana}=\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}=3\)

Chọn C.

Theo giả thiết ta có:

P = ( sina + sinb) ² + ( cosa + cosb) ²

= sin²a + 2.sina.sinb + sin²b + cos²a + 2cosa. cosb + cos²b

= 2 + 2( sina.sinb + cos a. cosb)

= 2 + 2.cos( a - b)   ( sử dụng công thức cộng)

\(2\cdot\left|-21\right|-3\cdot\left|125\right|-5\cdot\left|-33\right|-\left|2\cdot21\right|\)

\(=2\cdot21-3\cdot125-5\cdot33-2\cdot21\)

\(=-3\cdot125-5\cdot33=-375-165=-540\)

\(A=2x^2+2\sqrt{2}x+3\\ =2\left(x^2+\sqrt{2}x+\dfrac{3}{2}\right)\\ =2.\left(x^2+2.\dfrac{1}{\sqrt{2}}x+\dfrac{1}{2}+1\right)\\ =2.\left(x^2+2.\dfrac{1}{\sqrt{2}}x+\dfrac{1}{2}\right)+2\\ =2.\left(x+\dfrac{1}{\sqrt{2}}\right)^2+2\)

Ta có \(2.\left(x+\dfrac{1}{\sqrt{2}}\right)^2\ge0\forall x\)

\(2.\left(x+\dfrac{1}{\sqrt{2}}\right)^2+2\ge2\forall x\)

Dấu bằng xảy ra khi : \(x+\dfrac{1}{\sqrt{2}}=0\\ \Rightarrow x=\dfrac{-\sqrt{2}}{2}\)

Vậy \(Min_A=2\) khi \(x=\dfrac{-\sqrt{2}}{2}\)

16/5

17/4

a. 7/2 - 3/10 = 16/5

b. (4/6 + 5):4/3=17/3 :4/3=17/4

a) \(6-\dfrac{5}{2}-\dfrac{3}{10}=\dfrac{7}{2}-\dfrac{3}{10}=\dfrac{16}{5}\)

b) \(\dfrac{4}{6}\div\dfrac{4}{3}+5\div\dfrac{4}{3}\)

    \(=\dfrac{1}{2}+\dfrac{15}{4}\)

     \(=\dfrac{17}{4}\)