\(A=\dfrac{3}{2}-tana\cdot cos^2a\)

\(=\dfrac{3}{2}-\dfrac{sina}{cosa}\cdot cos^2a\)

\(=\dfrac{3}{2}-sina\cdot cosa\)

\(=\dfrac{3}{2}-\dfrac{1}{2}sin2a\)
\(0^0< a< 90^0\)

=>\(0< =2a< =180^0\)

=>\(sin2a\in\left[-1;1\right]\)

\(-1< =sin2a< =1\)

=>\(\dfrac{1}{2}>=-\dfrac{1}{2}sin2a>=-\dfrac{1}{2}\)

=>\(\dfrac{7}{2}>=-\dfrac{1}{2}sin2a+3>=\dfrac{5}{2}\)

=>\(\dfrac{5}{2}< =y< =\dfrac{7}{2}\)

\(y_{min}=\dfrac{5}{2}\) khi sin2a=1

=>\(2a=\dfrac{\Omega}{2}+k2\Omega\)

=>\(a=\dfrac{\Omega}{4}+k\Omega\)

mà 0<a<90

nên a=45

b a c a a a

lỗi

lỗi

Lỗi rồi bạn ơi

\(\dfrac{4}{7}+\dfrac{2}{9}+\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{7}{9}+\dfrac{75}{100}\\ =\dfrac{4}{7}+\dfrac{2}{9}+\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{7}{9}+\dfrac{3}{4}\\ =\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =\dfrac{7}{7}+\dfrac{9}{9}+\dfrac{4}{4}\\ =1+1+1\\ =3\)

Ta có: \(\dfrac{7}{11}=\dfrac{7\times3}{11\times3}=\dfrac{21}{33};\dfrac{8}{11}=\dfrac{8\times3}{11\times3}=\dfrac{24}{33}\)

2 phân số giữa là: \(\dfrac{22}{33};\dfrac{23}{33}\)

\(=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=1+1+1=3\)

(\(\dfrac{4}{7}\)+\(\dfrac{3}{7}\))+(\(\dfrac{2}{9}\)+\(\dfrac{7}{9}\))+(\(\dfrac{1}{4}\)+\(\dfrac{75}{100}\))

=1+1+1

=3

a) Áp dụng tính chất dãy tỉ số bằng nhau:

 \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y+z}{3+4+5}=\dfrac{360}{60}=6\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=6\\\dfrac{y}{4}=6\\\dfrac{z}{5}=6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=24\\z=30\end{matrix}\right.\)

b) Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{x}{-2}=-\dfrac{y}{4}=\dfrac{z}{5}=-\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{-2+8+15}=\dfrac{1200}{21}=\dfrac{400}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-2}=\dfrac{400}{7}\\-\dfrac{y}{4}=\dfrac{400}{7}\\\dfrac{z}{5}=\dfrac{400}{7}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{800}{7}\\y=-\dfrac{1600}{7}\\z=\dfrac{2000}{7}\end{matrix}\right.\)

c) Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}=\dfrac{-2z}{4}=\dfrac{x+y-2z}{5+1+4}=\dfrac{160}{10}=16\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=16\\\dfrac{y}{1}=16\\\dfrac{z}{-2}=16\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=80\\y=16\\z=-32\end{matrix}\right.\)