a: S

b: Đ

a) S

b) Đ

\(a,\dfrac{6\times2+4}{7\times2}=\dfrac{12+4}{14}=\dfrac{16}{14}=\dfrac{8}{7}\\ \Rightarrow S\\ b,\dfrac{6\times2+4}{7\times2+4}=\dfrac{16}{18}=\dfrac{8}{9}\\ \RightarrowĐ\)

1

đúng  vì : 4/7 : 4 = 4/7 : 4/1 = 4/28  mà 4/28 rút gọn = 1/7  nên đáp án là : đúng

\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

Biến đổi tử số 

\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)

= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)

= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)

= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)

Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)

Vậy A = 20

c.ơn nhìu a

\(a,\dfrac{11}{22}-\dfrac{3}{16}.\dfrac{8}{18}+\dfrac{1}{18}=\dfrac{1}{2}-\dfrac{1}{12}+\dfrac{1}{18}=\dfrac{18}{36}-\dfrac{3}{36}+\dfrac{2}{36}=\dfrac{17}{36}\)

\(b,\dfrac{5}{7}+\dfrac{3}{7}:4-\dfrac{8}{9}.\dfrac{-3}{4}=\dfrac{5}{7}+\dfrac{3}{7}.\dfrac{1}{4}-\dfrac{-2}{3}=\dfrac{5}{7}+\dfrac{3}{28}+\dfrac{2}{3}=\dfrac{60}{84}+\dfrac{9}{84}+\dfrac{56}{84}=\dfrac{125}{84}\)

Lời giải:

Không mất tổng quát giả sử $C$ là góc nhọn.

\(\sin ^2A+\sin ^2B+\sin ^2C=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\sin ^2C\)

\(=1+\sin ^2C-\frac{1}{2}(\cos 2A+\cos 2B)=1+\sin ^2C-\cos (A+B)\cos (A-B)\)

\(=1+\sin ^2C-\cos (180^0-C)\cos (A-B)\)

\(=1+\sin ^2C+\cos C\cos (A-B)=2-\cos ^2C+\cos C\cos (A-B)\)

\(\leq 2-\cos ^2C+\cos C\)  với mọi $C$ nhọn

\(=\frac{9}{4}-(\cos C-\frac{1}{2})^2\leq \frac{9}{4}\)

Do đó mệnh đề đã cho đúng.

Đúng

Sai

Sai

Đúng

Đúng

Sai

Đúng

Đúng

(Uy tín nha)