BC=2*5=10cm

Đặt BH=x; CH=y

Theo đề, ta có: xy=4^2=16 và x+y=10

=>(x,y)=(2;8) hoặc (x,y)=(8;2)

TH1: BH=2cm; CH=8cm

AB=căn 2*10=2*căn 5(cm)

AC=căn 8*10=4*căn 5(cm)

tan C=AB/AC=1/2

=>\(\widehat{C}\simeq27^0\)

=>\(\widehat{B}=63^0\)

TH2: BH=8cm; CH=2cm

AC=căn 2*10=2*căn 5(cm)

AB=căn 8*10=4*căn 5(cm)

tan B=AC/AB=1/2

=>\(\widehat{B}\simeq27^0;\widehat{C}=63^0\)

45D

\(B=\dfrac{x+3+2\left(\sqrt{x}-3\right)+\sqrt{x}+3}{x-9}\)

\(=\dfrac{x+\sqrt{x}+6+2\sqrt{x}-6}{x-9}=\dfrac{x+3\sqrt{x}}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

\(B=\dfrac{x+3}{x-9}+\dfrac{2}{3+\sqrt{x}}-\dfrac{1}{3-\sqrt{x}}\\ B=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\\ B=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x+3+2\sqrt{x}-6+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}\left(\text{đ}pcm\right)\)

a: \(=\dfrac{3\sqrt{5}+3-3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)}=\dfrac{4\sqrt{5}}{-2+2\sqrt{5}}=\dfrac{5+\sqrt{5}}{2}\)

b: \(=\dfrac{3\sqrt{2}+3-3+2\sqrt{2}}{\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}=\dfrac{5\sqrt{2}}{\sqrt{2}-1}=10+5\sqrt{2}\)

c: \(=\dfrac{14\sqrt{7}+28-18+6\sqrt{7}}{\left(3-\sqrt{7}\right)\left(\sqrt{7}+2\right)}\)

\(=\dfrac{20\sqrt{7}+10}{-1+\sqrt{7}}=25+5\sqrt{7}\)

d: \(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

\(\sqrt{28-16\sqrt{3}}-\sqrt{\left(4-3\sqrt{3}\right)^2}\)

\(=\sqrt{16-16\sqrt{3}+12}-\left|4-3\sqrt{3}\right|\)

\(=\sqrt{4^2-2.4.2\sqrt{3}+\left(2\sqrt{3}\right)^2}-\left(3\sqrt{3}-4\right)\)

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

\(=\left|4-2\sqrt{3}\right|-3\sqrt{3}+4\)

\(=4-2\sqrt{3}-3\sqrt{3}+4\)

\(=8-5\sqrt{3}\).

\(=4-2\sqrt{3}+3\sqrt{3}-4=\sqrt{3}\)

\(1,\sqrt{6+2\sqrt{5}}=\sqrt{\sqrt{5^2}+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}+1\right|=\sqrt{5}+1\\ 3,\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\\ =\sqrt{\sqrt{6^2}-2.3\sqrt{6}+3^2}-\sqrt{\sqrt{6^2}-2.2\sqrt{6}+2^2}\\ =\sqrt{\left(\sqrt{6}-3\right)^2}-\sqrt{ \left(\sqrt{6}-2\right)^2}\\ =\left|\sqrt{6}-3\right|-\left|\sqrt{6}-2\right|\\ =3-\sqrt{6}-\sqrt{6}+2\\ =-2\sqrt{6}+5\)

\(5,\sqrt{31-10\sqrt{6}}-\sqrt{\left(3-2\sqrt{6}\right)^2}\\ =\sqrt{\sqrt{6^2}-2.5\sqrt{6}+5^2}-\left|3-2\sqrt{6}\right|\\ =\sqrt{\left(\sqrt{6}-5\right)^2}-\left(2\sqrt{6}-3\right)\\ =\left|\sqrt{6}-5\right|-2\sqrt{6}+3\\ =5-\sqrt{6}-2\sqrt{6}+3\\ =8-3\sqrt{6}\)

Không đăng lại câu hỏi c nhé.

a)

\(C=\dfrac{3}{\sqrt{2}-1}-\dfrac{14}{3+\sqrt{2}}\\ =\dfrac{3\left(\sqrt{2}+1\right)}{2-1}-\dfrac{14\left(3-\sqrt{2}\right)}{9-2}\\ =3\sqrt{2}+3-\dfrac{14\left(3-\sqrt{2}\right)}{7}\\ =3\sqrt{2}+3-2\left(3-\sqrt{2}\right)\\ =3\sqrt{2}+3-6+2\sqrt{2}\\ =5\sqrt{2}-3\)

b)

\(\dfrac{4}{\sqrt{11}-3}-\dfrac{7}{2+\sqrt{11}}\\ =\dfrac{4\left(\sqrt{11}+3\right)}{11-9}-\dfrac{7\left(2-\sqrt{11}\right)}{4-11}\\ =\dfrac{4\left(\sqrt{11}+3\right)}{2}-\dfrac{7\left(2-\sqrt{11}\right)}{-7}\\ =2\left(\sqrt{11}+3\right)+2-\sqrt{11}\\ =2\sqrt{11}+6+2-\sqrt{11}\\ =\sqrt{11}+8\)

c)

\(B=\dfrac{1}{\sqrt{5}-2}-\dfrac{8}{\sqrt{5}+1}\\ =\dfrac{\sqrt{5}+2}{5-4}-\dfrac{8\left(\sqrt{5}-1\right)}{5-1}\\ =\sqrt{5}+2-\dfrac{8\left(\sqrt{5}-1\right)}{4}\\ =\sqrt{5}+2-2\left(\sqrt{5}-1\right)\\ =\sqrt{5}+2-2\sqrt{5}+2\\ =-\sqrt{5}+4\)

d)

\(M=\dfrac{11}{4-\sqrt{5}}-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\\ =\dfrac{11\left(4+\sqrt{5}\right)}{16-5}-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\\ =\dfrac{11\left(4+\sqrt{5}\right)}{11}-\sqrt{5}\\ =4+\sqrt{5}-\sqrt{5}\\ =4\)