A B C H 50 37 O O

Kẻ \(AH\perp BC\). Đặt BH = x thì \(CH=60-x\)

Xét tam giác vuông ABH có: \(AH=tan50^o.x\)

Xét tam giác vuông ACH có: \(AH=tan37^o.\left(60-x\right)\)

Vậy nên ta có: \(tan50.x=tan37^o.\left(60-x\right)\)

\(\Leftrightarrow\left(tan50^o+tan37^o\right).x=tan37^o.60\)

\(\Leftrightarrow x=\frac{tan37^o.60}{tan50^o+tan37^o}\)  (cm)

Vậy thì \(AB=\frac{x}{cos50^o}=\frac{tan37^o.60}{cos50^o\left(tan50^o+tan37^o\right)}\)  (cm)

\(AH=x.tan50^o=\frac{tan50^o.tan37^o.60}{\left(tan50^o+tan37^o\right)}\)  (cm)

\(AC=\frac{AH}{sin37^o}=\frac{tan50^o.60}{cos37^o\left(tan50^o+tan37^o\right)}\)  (cm)

\(S_{ABC}=\frac{1}{2}.BC.AH=\frac{30tan50^o.tan37^o.60}{tan50^o+tan37^o}=\frac{1800tan50^o.tan37^o}{tan50^o+tan37^o}\left(cm^2\right)\)

Bài 2:

\(x^2-2x-3-\left(x+1\right)\sqrt{x^2+3}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)-\left(x+1\right)\sqrt{x^2+3}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3-\sqrt{x^2+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3-\sqrt{x^2+3}=0\end{cases}}\)

TH1: \(x+1=0\Leftrightarrow x=-1\) 

TH2:  \(x-3-\sqrt{x^2+3}=0\Leftrightarrow x-3=\sqrt{x^2+3}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge3\\x^2+3=x^2-6x+9\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x=1\end{cases}}\left(l\right)\)

Vậy phương trình có nghiệm x =  -1.

ta có:2(y+z)=x(yz-1)

=>2y+2z=xyz-x

=>2y+2z+x=xyz

mik ko làm tiếp đc do thiếu đ/k

làm xong rồi thì please_sign

áp dụng bđt huyền thoại \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\) =\(\frac{a+b+c}{abc}=\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)}\) 

mà \(\left(ab+bc+ac\right)^2\ge3abc\left(a+b+c\right)\) (tụ cm nhé )

\(\Rightarrow\ge\frac{\left(a+b+c^2\right)}{\frac{\left(ab+bc+ac\right)^2}{3}}=\frac{3\left(a+b+c\right)^2\left(a^2+b^2+c^2\right)}{\left(ab+bc+ac\right)^2\left(a^2+b^2+c^2\right)}\)

m,à \(\left(ab+bc+ac\right)^2\left(a^2+b^2+c^2\right)\le\frac{\left(a^2+b^2+c^2+ab+bc+ac+ab+bc+ac\right)^3}{3^3}\)

   =\(\frac{\left(\left(a+b+c\right)^2\right)^3}{27}=27\)

\(\Rightarrow vt\ge\frac{27\left(a^2+b^2+c^2\right)}{27}=a^2+b^2+c^2\)

dau = khi a=b=c=1

hay quá bạn ơi

Hình như đề sai rồi bạn ơi. thử thay a=1,5 b=1 c=0,5 xem