\(\frac{x+\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}.\)

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

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

\(a,\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

\(=\sqrt{13+6\sqrt{4+\sqrt{1-2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}}\)

\(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)

\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

\(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+6\sqrt{1+2\sqrt{2}+2}}\)

\(=\sqrt{13+6\sqrt{\left(1+\sqrt{2}\right)^2}}\)

\(=\sqrt{13+6\left(1+\sqrt{2}\right)}=\sqrt{13+6+\sqrt{12}}\)

\(=\sqrt{19+2\sqrt{3}}\)

a) = \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

=    \(\sqrt{13+6\sqrt{4+\sqrt{8-2.2\sqrt{2}+1}}}\)

=    \(\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)

=    \(\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

=     \(\sqrt{13+6\sqrt{2+2\sqrt{2}+1}}\)

=     \(\sqrt{13+6\left(\sqrt{2}+1\right)}\)

=     \(\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}\)

=      \(\sqrt{18+2.3\sqrt{2}+1}\)

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

=       \(3\sqrt{2}+1\)

<=>   \(3x^2+3x=2\sqrt{x^2+x}+1\)

<=>    Đặt   \(\sqrt{x^2+x}=a\) \(\left(a\ge0\right)\)ta có

    \(3a^2-2a-1=0\)

có a+b+c=0

=> a₁=1    => \(x^2+x=1\)

=> a₂=-1/3  => x²+x=-1/3

dùng công thức nghiệm tính nốt nha bn! :D

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

=>\(\left(x+\sqrt{x-1}\right)\left(4x-4\sqrt{x-1}\right)=0\)

từ đay tự giải nốt

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

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

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

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

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

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

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

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)

_Minh ngụy_

\(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\) Đk \(x\ne0\)

\(=\frac{\sqrt{x^4-6x^2+9+12x^2}}{\sqrt{x^2}}+\sqrt{x^2+4x+4-8x}\)

\(=\frac{\sqrt{x^4+6x^2+9}}{\sqrt{x^2}}+\sqrt{x^2-4x+4}\)

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

\(=\frac{x^2+3}{x}+x-2\)

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

\(=\frac{2x^2-2x+3}{x}\)

b, \(P=\frac{2x^2-2x+3}{x}=2x-2+\frac{3}{x}\)

Để \(P\in z\)thì \(x\inƯ\left(3\right)=\left(-3;-1;1;3\right)\)

Sửa chút đề nhé , \(\frac{AB}{BC}=\frac{5}{6}\)Mới là đề đúng nha bạn, Giải :

Ta có  \(\Delta ABC\)vuông tại B 

\(\Rightarrow AB^2=AH.AC\)\(\Rightarrow AH=\frac{AB^2}{AC}\)\(\left(1\right)\)

\(BC^2=HC.AC\)\(\Rightarrow HC=\frac{BC^2}{AC}\)\(\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{AH}{HC}=\frac{AB^2}{AC}\div\frac{BC^2}{AC}=\frac{AB^2.AC}{AC.BC^2}=\frac{AB^2}{BC^2}=\frac{5^2}{6^2}=\frac{25}{36}\)

Đặt \(\frac{AH}{HC}=\frac{25}{36}=x\)\(\Rightarrow AH=25x;HC=36x\)

Mà \(BH^2=AH.HC=25x.36x=30^2=900\)

\(\Rightarrow x^2=\frac{900}{25.36}=\frac{900}{900}=1\)\(\Rightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{cases}}\)

\(\Rightarrow AH=25x=25.1=25\)

\(HC=36.x=36.1=36\)

\(KL:AH=25;HC=36\)

\(A^3=14+3\sqrt[3]{\left(7-\sqrt{50}\right)\left(7+\sqrt{50}\right)}\left(\sqrt[3]{7-\sqrt{50}}+\sqrt[3]{7+\sqrt{50}}\right)\)

\(A^3=14+3\sqrt[3]{49-50}.A\)\(\Leftrightarrow\)\(A^3=14-3A\)

\(\Leftrightarrow\)\(A^3+3A-14=0\)\(\Leftrightarrow\)\(A\left(A^2-4\right)+7\left(A-2\right)=0\)

\(\Leftrightarrow\)\(A\left(A-2\right)\left(A+2\right)+7\left(A-2\right)=0\)

\(\Leftrightarrow\)\(\left(A-2\right)\left(A^2+2A+7\right)=0\)

\(\Leftrightarrow\)\(A=2\) ( do \(A^2+2A+7=\left(A+1\right)^2+6>0\) )