Cho x,y >0 thỏa mãn \(x+y\le2\). Tìm GTNN của :

\(A=x+y+\frac{2}{x}+\frac{2}{y}\)

Các bạn giúp mk nhé !

\(\sqrt{\left(a-b\right)^2}+\sqrt{\left(a+b\right)^2}\)

\(=|a-b|+|a+b|\) ( 1 )

Nếu \(a\ge b\) \(\Rightarrow\) \(|a-b|=a-b\)

( 1 ) \(\Rightarrow\) a - b + a + b = 2a

Nếu \(a\le b\) \(\Rightarrow\) \(|a-b|=b-a\)

( 1 ) \(\Rightarrow\) b - a + a + b = 2b

2x² + 3x + 1 = 0

\(\Rightarrow\) 2x² + 2x + x + 1 = 0

\(\Rightarrow\) 2x ( x + 1 ) + ( x + 1 ) = 0

\(\Rightarrow\) ( x + 1 )( 2x + 1 ) = 0

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\) \(S=\left\{-1;-\frac{1}{2}\right\}\)

( x² + 1 )( x - 1 ) = 0

Mà x² + 1 > 0

=> x - 1 = 0

=> x = 1

a, Xét \(\Delta BHK\) và \(\Delta BAD\) có :

\(\widehat{B_2}=\widehat{B_1}\left(gt\right)\)

\(\widehat{BHK}=\widehat{BAD}=90^o\)

\(\Rightarrow\) \(\Delta BHK\sim\Delta BAD\left(g.g\right)\)

Xét \(\Delta BAK\) và \(\Delta BCD\) có :

\(\widehat{B_1}=\widehat{B_2}\left(gt\right)\)

\(\widehat{BAK}=\widehat{BCD}\) ( cùng phụ với \(\widehat{ABC}\) )

\(\Rightarrow\) \(\Delta BAK\sim\Delta BCD\left(g.g\right)\)

b, Ta có : \(\Delta BHK\sim\Delta BAD\) ( câu a )

\(\Rightarrow\) \(\frac{HK}{AD}=\frac{HB}{BA}\)

Mà BK là phân giác \(\widehat{ABH}\)

\(\Rightarrow\) \(\frac{HB}{BA}=\frac{HK}{AK}\)

\(\Rightarrow\) \(\frac{HK}{AD}=\frac{HK}{AK}\) \(\Rightarrow\) \(AD=AK\)

Lại có : \(\Delta BAK\sim\Delta BCD\) ( câu a )

\(\Rightarrow\) \(\frac{AK}{CD}=\frac{BK}{BD}\)

Mà \(\frac{BK}{BD}=\frac{HK}{AD}\left(\Delta BHK\sim\Delta BAD\right)\)

\(\Rightarrow\) \(\frac{AK}{CD}=\frac{HK}{AD}\)

\(\Rightarrow\) \(AK.AD=HK.DC\) Mặt khác : AD = AK
\(\Rightarrow\) \(AK^2=HK.DC\)

a, Xét \(\Delta ACF\) và \(\Delta HCE\) có :

\(\widehat{C_1}=\widehat{C_2}\left(gt\right)\)

\(\widehat{FAC}=\widehat{EHC}=90^o\)

\(\Rightarrow\) \(\Delta ACF\sim\Delta HCE\left(g.g\right)\)

a, Xét \(\Delta ABC\) và \(\Delta HBA\) có :

\(\widehat{B}:chung\)

\(\widehat{BAC}=\widehat{AHB}=90^o\)

\(\Rightarrow\) \(\Delta ABC\sim\Delta HBA\left(g.g\right)\)

b, Xét \(\Delta AHB\) và \(\Delta DHC\) có :

\(\widehat{AHB}=\widehat{DHC}=90^o\)

\(\widehat{BAH}=\widehat{HDC}\) ( so le trong )

\(\Rightarrow\) \(\Delta AHB\sim\Delta DHC\left(g.g\right)\)