Đặt \(d=\left(a,b\right)\)

Suy ra \(a=dm,b=dn,\left(m,n\right)=1\).

\(a^2+b^2=d^2\left(m^2+n^2\right)\)

\(ab=d^2mn\)

Suy ra \(\left(m^2+n^2\right)⋮mn\Rightarrow\hept{\begin{cases}m^2+n^2⋮m\\m^2+n^2⋮n\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}m^2⋮n\\n^2⋮m\end{cases}}\Leftrightarrow\hept{\begin{cases}m⋮n\\n⋮m\end{cases}}\)(vì \(\left(m,n\right)=1\))

Suy ra \(m=n=1\).

Do đó \(a=b\)

\(M=\frac{8ab}{a^2+b^2}=\frac{8a^2}{a^2+a^2}=4\)là số chính phương. 

Sai thông cảm ạ

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

\(A=\frac{x-1}{\sqrt{x}-2}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-2}=\frac{\left(\sqrt{3}-1+1\right)\left(\sqrt{3}-1-1\right)}{\sqrt{3}-1-2}=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-3}\)

\(=\frac{\sqrt{3}-2}{1-\sqrt{3}}\)

em mới lớp 7, nếu em làm sai mong chị thông cảm nha!

Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}-\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(A^2=4+\sqrt{10+2\sqrt{5}}-2\sqrt{16-\left(10+2\sqrt{5}\right)}+4-\sqrt{10+2\sqrt{5}}\)

\(=8-2\sqrt{6-2\sqrt{5}}=8-2\sqrt{\left(\sqrt{5}-1\right)^2}=8-2\left(\sqrt{5}-1\right)\)

\(=8-2\sqrt{5}+2=10-2\sqrt{5}\)

Vậy \(A=\sqrt{10-2\sqrt{5}}\)

chịu khó thế

78, Với \(x>0;x\ne1\)

\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

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

79, Với a > 0 ; \(a\ne1\)

\(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)

\(=\left(\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(=\left(\frac{4\sqrt{a}+4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)=\frac{4a\sqrt{a}\left(a+1\right)}{\sqrt{a}\left(a-1\right)}=\frac{4a\left(a+1\right)}{a-1}\)

a, Theo định lí Pytago tam giác ABH : \(AB^2=AH^2+BH^2\)(1)

Theo định lí Pytago tam giác ACH : \(AC^2=AH^2+AC^2\)(2) 

Lấy (1) - (2) : \(AB^2-AC^2=AH^2+BH^2-AH^2-HC^2\)

\(\Leftrightarrow AB^2-AC^2=BH^2-HC^2\Leftrightarrow AB^2+HC^2=BH^2+AC^2\)

b, Ta có : \(AH^2=AM.AB\)( hệ thức lượng ) (1) 

\(AH^2=AN.AC\)( hệ thức lượng ) (2) 

Từ (1) ; (2) suy ra : \(AM.AB=AN.AC\)(3) 

(3) => \(\frac{AM}{AC}=\frac{AN}{AB}\)

Xét tam giác AMN và tam giác ACB ta có : 

^A _ chung 

\(\frac{AM}{AC}=\frac{AN}{AB}\)

Vậy tam giác AMN ~ tam giác ACB ( c.g.c ) 

\(M=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)

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

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

\(=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)

\(M< \frac{3}{2}\Leftrightarrow\frac{2\sqrt{x}+1}{\sqrt{x}+1}< \frac{3}{2}\Leftrightarrow2\left(2\sqrt{x}+1\right)< 3\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow4\sqrt{x}+2< 3\sqrt{x}+3\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\).

\(\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\)

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

\(\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\Rightarrow0\le x< 9\)

anh đi anh nhớ quê nha 

nhớ canh rau muống nhớ cà dầm tương 

nhớ thằng đẩy bố xuống mương 

bố mà bắt được bố tương vỡ mồm