trên hình kẻ đường cao AH

ta có \(\text{cos 50}=\frac{HC}{AC}\Rightarrow HC=AC.\text{cos 50 }=35.\text{cos 50}\approx22\text{ (cm)}\)

\(\text{sin 50}=\frac{AH}{AC}\Rightarrow AH=AC.\text{sin 50}=35.\text{sin 50}\approx27\left(cm\right)\)

\(\text{tan 60}=\frac{AH}{BH}\Rightarrow BH=\frac{AH}{\text{tan 60}}=\frac{27}{\text{tan 60}}\approx16\left(cm\right)\)

\(\Rightarrow BC=22+16=38\left(cm\right)\)

\(\text{sin 60}=\frac{AH}{AB}\Rightarrow AB=\frac{AH}{\text{sin 60}}=\frac{27}{\text{sin 60}}\approx31\left(cm\right)\)

Diện tích tam giác ABC là:

35+38+31=104 (cm)

\(a)\sqrt{-9a}-\sqrt{9+12a+4a^2}\)

\(==\sqrt{3^2.\left(-a\right)}-\sqrt{3^2-2.3.2a+\left(2a\right)^2}\)

\(=3\sqrt{-a}-\sqrt{\left(3+2a\right)^2}\)

\(=3\sqrt{a}-\left|3+2a\right|\)

\(b)1+\frac{3m}{m-2}\sqrt{m^2-4m+4}\)

\(=1+\frac{3m}{m-2}\sqrt{\left(m\right)^2-2.2m+2^2}\)

\(=1+\frac{3m}{m-2}\sqrt{\left(m-2\right)^2}\)

\(=1+\frac{3m}{m-2}|m-2|\)

\(c)4x-\sqrt{9x^2+6x+1}\)

\(=4x-\sqrt{\left(3x\right)^2+2.3x+1}\)

\(=4x-\sqrt{\left(3x+1\right)^2}\)

\(=4x-|3x+1|\)

Câu hỏi của Nguyễn Đa Vít - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo thêm!

\(a,x-9+y-2\sqrt{xy}\left(x;y>0\right)\)

\(=\left(\sqrt{x}\right)^2-2\sqrt{x}\sqrt{y}+\left(\sqrt{y}\right)^2-9\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2-9\)

\(=\left(\sqrt{x}-\sqrt{y}+3\right)\left(\sqrt{x}-\sqrt{y}-3\right)\)

\(b,\text{ đkxđ }x\ge0\)

\(x-5\sqrt{x}+6=\left(\sqrt{x}\right)^2-2\sqrt{x}-3\sqrt{x}+6\)

\(=\sqrt{x}.\left(\sqrt{x}-2\right)-3.\left(\sqrt{x}-2\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

\(c,đ\text{kxđ }x\ge0\)

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

\(=\sqrt{x}\left(\sqrt{x}+1\right)+3.\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)

\(d,\text{đkxđ }x\ge0\)

\(\sqrt{x}-x^2=\sqrt{x}-\left(\sqrt{x}\right)^4=\sqrt{x}\left(1-\left(\sqrt{x}\right)^3\right)\)

\(=\sqrt{x}.\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)\)

\(B=2\sqrt{2016}=\sqrt{2016}+\sqrt{2016}>\sqrt{2014}+\sqrt{2016}=A\)

 Vậy A < B

ta có \(B=2\sqrt{2016}=\sqrt{2016}+\sqrt{2016}>\sqrt{2014}+\sqrt{2016}=A\)

Vậy A<B