Đáp án A

M = 5 6 : 5 2 2 + 7 15 M =   5 6 : 25 4 + 7 15 M =   5 6 . 4 25 + 7 15 M =   1 . 2 3 . 5 + 7 15 M =   2 15 + 7 15 M =   9 15 = 3 5

Khi đó a = 3 , b = 5 nên a + b = 8

Đáp án A

M = 5 6 : 5 2 2 + 7 15 M =   5 6 : 25 4 + 7 15 M =   5 6 . 4 25 + 7 15 M =   1 . 2 3 . 5 + 7 15 M =   2 15 + 7 15 M =   9 15 = 3 5

Khi đó a = 3 , b = 5 nên a + b = 8

lỗi số r

\(cot\alpha=\dfrac{1}{2}tacó:\)

\(cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)

\(A=\dfrac{4sin\alpha+5cos\alpha}{2sin\alpha-3cos\alpha}\)

\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)

\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)

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

\(A=13\)

20x30 x...100 =  có 10 chữ số 0

A có 8 chữ số hàng đơn vị là 5

Số hàng đơn vị 5 nhân với số chẵn thêm 1 số 0, riêng 25 x 24 có 2 số 0

Tổng có 19 số 0

\(A=\dfrac{4\sin\alpha+5\cos\alpha}{2\sin\alpha-3\cos\alpha}\)

\(A=\dfrac{\dfrac{4\sin\alpha}{\sin\alpha}+\dfrac{5\cos\alpha}{\sin\alpha}}{\dfrac{2\sin\alpha}{\sin\alpha}-\dfrac{3\cos\alpha}{\sin\alpha}}\)

\(A=\dfrac{4+5\cot\alpha}{2-3\cot\alpha}\)

Thay cot α= \(\dfrac{1}{2}\)   vào A, ta có:

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

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

\(A=\dfrac{13}{\dfrac{2}{\dfrac{1}{2}}}\)

A=13

\(A=\dfrac{\dfrac{4sin\alpha}{sin\alpha}+\dfrac{5cos\alpha}{sin\alpha}}{\dfrac{2sin\alpha}{sin\alpha}-\dfrac{3cos\alpha}{sin\alpha}}\)

\(A=\dfrac{4+5cot\alpha}{2-3cot\alpha}\)

Biết cotα=\(\dfrac{1}{2}\) nên ta có:

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

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

A= 13

\(\dfrac{3sin\alpha-4cos\alpha}{2sin\alpha+3cos\alpha}=\dfrac{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}{\dfrac{2sin\alpha}{cos\alpha}+\dfrac{3cos\alpha}{cos\alpha}}=\dfrac{3tan\alpha-4}{2tan\alpha+3}\)

Biết tanα=\(-\dfrac{1}{4}\) nên ta có:

\(\dfrac{3\cdot\dfrac{-1}{4}-4}{2\cdot\dfrac{-1}{4}+3}=\dfrac{-\dfrac{3}{4}-4}{-\dfrac{1}{2}+3}=\dfrac{-19}{10}\)