nhân VT ra chuyển vế bình tiếp

bạn tự vẽ hình nha

áp dụng hệ thức lượng trong tam giác vuông ABCco \(AB^2=BA'^2\cdot BC,AC^2=A'C^2\cdot BC\)

                                                                                     \(\Rightarrow\frac{AB^2}{AC^2}=\frac{BA'}{A'C}\Rightarrow\frac{AC^4}{AB^4}=\frac{A'C^2}{A'B^2}\) (1) 

mà trong tam giác vuông AA'B có\(BA'^2=BF\cdot AB\)

 trong tam giác vuông AA'C có \(A'C^2=EC\cdot AC\)  

thay vào (1) ta co \(\frac{AC^4}{AB^4}=\frac{EC\cdot AC}{BF\cdot AB}\Rightarrow\frac{AC^3}{AB^3}=\frac{EC}{BF}\left(DPCM\right)\) 

b,de dang chung minh duoc tam giac BMD~BAC 

SUY RA \(\frac{BD}{BC}=\frac{BM}{BA}=\frac{MD}{AC}\) (2)

tuong tu tam giac NDC~ABC 

SUY RA \(\frac{DC}{BC}=\frac{NC}{AC}=\frac{ND}{AB}\)(3)

nhan (2) voi (3) ta co \(\frac{BD\cdot DC}{BC^2}=\frac{BM\cdot ND}{AB^2}=\frac{MD\cdot NC}{AC^2}=\frac{BM\cdot ND+MD\cdot NC}{AB^2+AC^2}\)

suy ra \(BD\cdot DC=BM\cdot ND+MD\cdot NC\) 

de dang cm duoc tu giac AMDN  la hcn suy ra MA =ND,MD=AN

THAY VAO BIEU THUC TREN TA CO \(BD\cdot DC=MA\cdot MB+NA\cdot NC\left(DPCM\right)\)

\(x^2-x-1=x^2-x+\frac{1}{4}-\frac{5}{4}=\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2=\left(x-\frac{1-\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)\)

mk chưa học đến lớp 9 nên mk ko biết !

a,

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

b,

A=\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+2\sqrt{12}}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-1-\sqrt{12}}}}{\sqrt{6}+\sqrt{2}}\)\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)

B=

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

A=\(\left(1-\sqrt{7}\right)x\frac{\sqrt{7}\left(1+\sqrt{7}\right)}{2\sqrt{7}}\)

=\(\frac{\left(1-7\right)\sqrt{7}}{2\sqrt{7}}\)

=-3

P=\(\frac{1+\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)}.\frac{-\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)}{\sqrt{x}}\)(ĐKXĐ:X>0,X KHÁC 1)

=\(\frac{-2\sqrt{x}}{\sqrt{x}}\)

=-2