\(VT=|2x+3|+|2x-1|=|2x+3|+|1-2x|\ge|2x+3+1-2x|=4\)

\(VP=\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)

Dấu \(=\)xảy ra khi \(x=-1\). 

mk quên mất cách giải nhưng 

có lời giải trên mạng r nhé

Vì ∆ABC vuông tại A nên : \(\widehat{B}\)+\(\widehat{C}\)= 90°

                                      => \(\widehat{C}\)= 90° - \(\widehat{B}\)= 90° - 50° = 40°

Vì ∆HCA vuông tại H nên : \(\widehat{C}\)+ \(\widehat{HAC}\)= 90°

                                      => \(\widehat{HAC}\)= 90° - \(\widehat{C}\)= 90° - 40° = 50°

Vậy \(\widehat{HAC}\)= 50°

(Hình tự vẽ)
Xét tam giác ABC có góc BAC=90^o 
=> Góc B+Góc C=90^o
=> 50^o+góc C=90^o
=> Góc C= 40^o
Xét tam giác AHC có góc AHC=90^o
=> Góc HAC+ Góc C=90^o
=> Góc HAC+ 40^o =90^o
=> Góc HAC= 50^o
 

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)

=>\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\right)=\frac{49}{99}\)

=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\right)=\frac{49}{99}\)

=> \(\frac{1}{2}\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)

=> \(1-\frac{1}{2x+1}=\frac{98}{99}\)

=> \(\frac{1}{2x+1}=\frac{1}{99}\)

=>2x + 1 = 99

=> 2x = 98

=> x = 49

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)

=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{49}{99}\)

=> \(1-\frac{1}{2x+1}=\frac{49}{99}\)

=> \(\frac{1}{2x+1}=1-\frac{49}{99}\)

=> \(\frac{1}{2x+1}=\frac{50}{99}\)

=> \(2x+1=\frac{99}{50}\)

=> \(2x=\frac{99}{50}-1\)

=> \(2x=\frac{49}{50}\)

=> \(x=\frac{49}{50}:2\)

=> \(x=\frac{49}{100}\)