\(A=\frac{1}{a^2+b^2}+\frac{25}{ab}+ab\)

\(A=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{16}{ab}+ab+\frac{17}{2ab}\)

áp dụng :  \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{\left(a+b\right)^2}\) mà  \(a+b\le4\)

\(\Rightarrow\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{1}{4}\)

theo cô si ta  có : \(\frac{16}{ab}+ab\Rightarrow2\sqrt{\frac{16}{ab}\cdot ab}=8\)

có \(ab\le\frac{\left(a+b\right)^2}{4}\le4\) \(\Rightarrow\frac{17}{2ab}\ge\frac{17}{8}\)

\(\Rightarrow A\ge\frac{17}{8}+8+\frac{1}{4}=\frac{83}{8}\)

dấu = xảy ra khi a=b=2

Ta có : \(cos^215^o=sin^275^o;cos^225^o=sin^265^o;cos^235^o=sin^255^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)

Khi đó \(N=sin^275^o+cos^275^o-\left(sin^265^o+cos^265^o\right)+sin^255^o+cos^255^o-\left(\frac{sin^245^0+cos^245^o}{2}\right)\)

Áp dụng công thức \(sin^2a+cos^2a=1\)ta được 

\(N=1-1+1-\frac{1}{2}=\frac{1}{2}\)

Vậy N = 1/2 

câu b chờ chút mình làm cho nhé <33

Ta có : \(cos^21^o=sin^289^o;cos^22^o=sin^288^o;...;cos^244^o=sin^246^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)

Khi đó \(A=\frac{sin^245^o+cos^245^o}{2}+\left(sin^246^0+cos^246^o\right)+...+\left(sin^289^o+cos^289^o\right)\)

Áp dụng ct \(sin^2a+cos^2a=1\)ta được \(A=\frac{1}{2}+1+1+...+1=...\)

P/S : bạn tự đếm xem bao nhiêu cặp nhé ;) tìm ssh á 

\(2\left|2x-\frac{5}{7}\right|-1\ge-1\)

Dấu ''='' xảy ra khi \(x=\frac{5}{7}:2=\frac{5}{14}\)

Vậy GTNN của biểu thức trên bằng -1 tại x = 5/14

\(2.\left|2x-\frac{5}{7}\right|-1\ge-1\)

dấu "=" xảy ra khi và chỉ khi \(2x-\frac{5}{7}=0< =>x=\frac{5}{14}\)

vậy \(MIN=-1\)

\(2\left|2x-\frac{5}{7}\right|-1\)

Vì \(2\left|2x-\frac{5}{7}\right|\ge0\forall x\)

\(\Rightarrow2\left|2x-\frac{5}{7}\right|-1\ge-1\forall x\)

Vậy  \(2\left|2x-\frac{5}{7}\right|-1\) đạt giá trị nhỏ nhất là \(-1\Leftrightarrow2x-\frac{5}{7}=0\Leftrightarrow2x=\frac{5}{7}\Leftrightarrow x=\frac{5}{14}\)

\(\frac{\cos a+\sin a}{\cos a-\sin a}=\frac{1+\tan a}{1-\tan a}=\frac{1+\frac{1}{7}}{1-\frac{1}{7}}=\frac{\frac{8}{7}}{\frac{6}{7}}=\frac{4}{3}\)

a) Ta có: sin30=cos60, sin50=cos40

    Mà cos30 < cos38 < cos40 < cos60 < cos80

    Nên cos30 < cos38 < sin50 < sin30 < cos80

b) Ta có: tan75=cot15, tan63=cot27 => cot11 < tan75 < cot20 < tan63 (1)

         và: sin49=cos41 => cos30 < sin49 (2)

    Lại có: cot11=tan69 > tan49= sin49:cos49 > sin49 (do cos49<1) (3)

    Từ (1), (2) và (3) suy ra: cos30 < sin49 < cot11 < tan75 < cot20 < tan63

TA CÓ   \(\sin30\)= \(\cos60\)

             \(\sin50=\cos40\)

---->>  \(\cos30< \cos38< \cos40< \cos60< \cos80\)

------>> \(\cos30< \cos38< \sin50< \sin60< \cos80\)

Cái kia làm tương tự nhoa

Bạn xin 1 cái k

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\frac{B}{A}< \frac{-1}{3}\Leftrightarrow\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}+\frac{1}{3}< 0\Leftrightarrow\frac{-9}{3\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{3\left(\sqrt{x}-3\right)}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-12}{3\left(\sqrt{x}-3\right)}< 0\). Xét hai trường hợp :

1.\(\hept{\begin{cases}\sqrt{x}-12>0\\3\left(\sqrt{x}-3\right)< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>144\\x< 9\end{cases}}\left(loai\right)\)

2. \(\hept{\begin{cases}\sqrt{x}-12< 0\\3\left(\sqrt{x}-3\right)>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 144\\x>9\end{cases}}\Rightarrow9< x< 144\)

Vậy ... 

\(2,B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

3, \(\frac{B}{A}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(\frac{B}{A}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(\frac{B}{A}=\frac{-3}{\sqrt{x}+3}\) mà \(\frac{B}{A}< -\frac{1}{3}\)

\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}< -\frac{1}{3}\)

\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}+\frac{1}{3}< 0\)

\(\Leftrightarrow\frac{-9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}< 0\) mà \(3\left(\sqrt{x}+3\right)>0\)

\(\Leftrightarrow-12-\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}>-12\left(luondung\right)\)