Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{\cot a - \cos a}}{{{{\cos }^3}a}} = \dfrac{{\dfrac{{\cos a}}{{\sin a}} - \cos a}}{{{{\cos }^3}a}} = \dfrac{{\cos a.\left( {\dfrac{1}{{\sin a}} - 1} \right)}}{{{{\cos }^3}a}}\\
 = \dfrac{{\dfrac{1}{{\sin a}} - 1}}{{{{\cos }^2}a}} = \dfrac{{\dfrac{{1 - \sin a}}{{\sin a}}}}{{1 - {{\sin }^2}a}} = \dfrac{{1 - \sin a}}{{\sin a.\left( {1 - \sin a} \right)\left( {1 + \sin a} \right)}}\\
 = \dfrac{1}{{\sin a.\left( {1 + \sin a} \right)}}\\
b,\\
\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\sin }^4}x + {{\cos }^4}x - {{\sin }^2}x}}\\
 = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x.{{\cos }^2}x - {{\sin }^2}x}}\\
 = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{1^2} - 2{{\sin }^2}x.{{\cos }^2}x - {{\sin }^2}x}}\\
 = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\left( {1 - {{\sin }^2}x} \right) - 2{{\sin }^2}x.{{\cos }^2}x}}\\
 = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x - 2{{\sin }^2}x.{{\cos }^2}x}}\\
 = \dfrac{{\left( {1 - {{\sin }^2}x} \right) - {{\sin }^2}x}}{{{{\cos }^2}x\left( {1 - 2{{\sin }^2}x} \right)}}\\
 = \dfrac{{1 - 2{{\sin }^2}x}}{{{{\cos }^2}x.\left( {1 - 2{{\sin }^2}x} \right)}}\\
 = \dfrac{1}{{{{\cos }^2}x}}\\
 = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}}\\
 = \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 1\\
 = 1 + {\tan ^2}x
\end{array}\)

1h.

$\sqrt{x+1}+\sqrt{4-x}+\sqrt{(x+1)(4-x)}≥5$

ĐK: $\left\{ \begin{array}{l}x+1≥0\\4-x≥0\end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l}x≥-1\\x≤4\end{array} \right.$

$\Leftrightarrow -1≤x≤4$

Đặt $t=\sqrt{x+1}+\sqrt{4-x} (t≥0)$

$\Rightarrow t^2=x+1+2\sqrt{(x+1)(4-x)}+4-x$

$\Rightarrow t^2=5+2\sqrt{(x+1)(4-x)}$

$\Rightarrow \dfrac{t^2-5}{2}=\sqrt{(x+1)(4-x)}$

$pt \Rightarrow \dfrac{t^2-5}{2}+t-5≥0$

$\Leftrightarrow t^2+2t-15 ≥0$

$\Leftrightarrow t ≤ -5$ hay $t≥3$

So với ĐK: $\Rightarrow t≥3$

Xét $t≥3:$

$\Leftrightarrow 5+2\sqrt{(x+1)(4-x)}≥9$

$\Leftrightarrow \sqrt{-x^2+3x+4}≥2$

$\Leftrightarrow \left\{ \begin{array}{l}2≤0(sai)\\-x^2+3x+4≥0\end{array} \right.$ hay $ \left\{ \begin{array}{l}2≥0(ld)\\-x^2+3x≥0\end{array} \right. $

$\Leftrightarrow \left\{ \begin{array}{l}2≤0(sai)\\-1≤x≤4\end{array} \right. $ hay $ \left\{ \begin{array}{l}2≥0(ld)\\0≤x≤3\end{array} \right. $

So với ĐK:

$\Rightarrow 0≤x≤3$ thoả ycbt