Bài 23 trang 201 SGK Đại số 10 Nâng cao

LG a

\(\sqrt {{{\sin }^4}\alpha  + 4{{\cos }^2}\alpha }  \) \(+ \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha } \)

Phương pháp giải:

Sử dụng công thức \[{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\]

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sqrt {{{\sin }^4}\alpha  + 4{{\cos }^2}\alpha }  \cr&=\sqrt {{{\sin }^4}\alpha + 4(1 - {{\sin }^2}\alpha )} \cr & = \sqrt {{{\sin }^4}\alpha  - 4{{\sin }^2}\alpha  + 4} \cr &= \sqrt {{{(2 - {{\sin }^2}\alpha )}^2}} \cr 
& = 2 - {\sin ^2}\alpha \,\,\,\,\,\,\,\,\,({\sin ^2}\alpha \le 1) \cr 
& \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha } \cr &=\sqrt {{{\cos }^4}\alpha + 4(1 - {{\cos }^2\alpha})} \cr & = \sqrt {{{\cos }^4}\alpha  - 4{{\cos }^2}\alpha  + 4} \cr &= \sqrt {{{(2 - {{\cos }^2}\alpha )}^2}} \cr 
& = 2 - {\cos ^2}\alpha \,\,\,\,\,\,\,\,(\cos ^2\alpha \le 1) \cr} \)

Vậy:

\(\sqrt {{{\sin }^4}\alpha  + 4(1 - {{\sin }^2}\alpha )}  \) \(+ \sqrt {{{\cos }^4}\alpha  + 4{{\sin }^2}\alpha } \)

\( = 2 - {\sin ^2}\alpha  + 2 - {\cos ^2}\alpha \) \(= 4-(\sin ^2\alpha +\cos ^2\alpha ) =4 - 1= 3\)

LG b

\(2(si{n^6}\alpha {\rm{ }} + {\rm{ }}co{s^6}\alpha {\rm{ }}){\rm{ }}-{\rm{ }}3(co{s^4}\alpha {\rm{ }} + {\rm{ }}si{n^4}\alpha {\rm{ }})\)

Phương pháp giải:

Sử dụng các hằng đẳng thức:

\[\begin{array}{l}
{A^3} + {B^3} = {\left( {A + B} \right)^3} - 3AB\left( {A + B} \right)\\
{A^2} + {B^2} = {\left( {A + B} \right)^2} - 2AB
\end{array}\]

Lời giải chi tiết:

 Ta có:

\(si{n^6}\alpha {\rm{ }} + {\rm{ }}co{s^6}\alpha \) 

\( = {\rm{ }}(si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha )^3{\rm{ }}\) \(-{\rm{ }}3si{n^2}\alpha co{s^2}\alpha {\rm{ }}(si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha )\)

\( = {\rm{ }}1{\rm{ }}-{\rm{ }}3si{n^2}\alpha {\rm{ }}co{s^2}\alpha \)

\(co{s^4}\alpha {\rm{ }} + {\rm{ }}si{n^4}\alpha {\rm{ }}\) \( = {\rm{ }}{(co{s^2}\alpha {\rm{ }} + {\rm{ }}si{n^2}\alpha )^2}-{\rm{ }}2si{n^2}\alpha {\rm{ }}co{s^2}\alpha \)

\( = {\rm{ }}1{\rm{ }} - {\rm{ }}2si{n^2}\alpha {\rm{ }}co{s^2}\alpha \)

Suy ra: 

\(\eqalign{
& 2\left( {{{\sin }^6}\alpha + {{\cos }^6}\alpha } \right) - 3({\cos ^4}\alpha + {\sin ^4}\alpha ) \cr} \)

\(\begin{array}{l}
= 2\left( {1 - 3{{\sin }^2}\alpha {{\cos }^2}\alpha } \right) - 3\left( {1 - 2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right)\\
= 2 - 6{\sin ^2}\alpha {\cos ^2}\alpha - 3 + 6{\sin ^2}\alpha {\cos ^2}\alpha \\
= 2 - 3 = - 1
\end{array}\)

LG c

\({2 \over {\tan \alpha  - 1}} + {{\cot \alpha  + 1} \over {\cot \alpha  - 1}}\,\,\,\,(\tan \alpha  \ne 1)\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& {2 \over {\tan \alpha - 1}} + {{\cot \alpha + 1} \over {\cot \alpha - 1}}\,\,\,\, \cr&= {2 \over {{1 \over {\cot \alpha }} - 1}} + {{\cot \alpha + 1} \over {\cot \alpha - 1}} \cr 
&  = \frac{2}{{\frac{{1 - \cot \alpha }}{{\cot \alpha }}}} + \frac{{\cot \alpha  + 1}}{{\cot \alpha  - 1}}\cr &= {{2\cot \alpha } \over {1 - \cot \alpha }} + {{\cot \alpha + 1} \over {\cot \alpha - 1}}\cr & = \frac{{2\cot \alpha }}{{1 - \cot \alpha }} - \frac{{\cot \alpha  + 1}}{{1 - \cot \alpha }} \cr &= \frac{{2\cot \alpha  - \cot \alpha  - 1}}{{1 - \cot \alpha }}\cr & = {{\cot \alpha - 1} \over {1 - \cot \alpha }} = - 1 \cr} \)

Loigiaihay.com