a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

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

b) Áp dụng đẳng thức ở câu a: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\ge\left(ac+bd\right)^2\)

Dấu "=" xảy ra khi \(\left(ad-bc\right)^2=0\Leftrightarrow ad=bc\)

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Bài làm : 

a.

(ac + bd)² + (ad – bc)²

= a² c² + 2acbd + b² d² + a² d² - 2adbc + b² c²

= a² c² + b² d² + a² d² + b² c²

= ( a² c² + a² d² ) + ( b² d² + b² c² )

= a² ( c² + d² ) + b² ( c² + d² )

= ( a² + b² ) . ( c² + d² )

Vậy (ac + bd)² + (ad – bc)² = (a² + b²)(c² + d²)

b.

( a² + b² ) . ( c² + d² ) - ( ac + bd )²

= a² c² + a² d² + b² c² + b² d² - a² c²  - 2acbd - b² d²

= a² d² + b² c² - 2acbd

= ( ad )² - 2ad . bc + ( bc )²

= ( ad - bc )² \(\ge\)0

\(\Rightarrow\) (ac + bd)² ≤ (a² + b²)(c² + d²)

Vậy (ac + bd)² ≤ (a² + b²)(c² + d²)

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\left(2\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{43+30\sqrt{2}}\)

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

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

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{4+2\sqrt{3}}\)

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

\(\sqrt{15-\sqrt{216}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3+2\sqrt{6}\right)^2}\)

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

a)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(1+2\sqrt{2}\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}=\sqrt{13+30\sqrt{\left(1+\sqrt{2}\right)^2}}=\sqrt{13+30\left(1+\sqrt{2}\right)}\)

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

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

\(=\sqrt{6+2\sqrt{5-\left(1+2\sqrt{3}\right)}}=\sqrt{6+2\sqrt{\left(1-\sqrt{3}\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

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

c)\(\sqrt{15-\sqrt{216}}+\sqrt{33+12\sqrt{6}}=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}\)

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

d) Sửa đề nha, đề bị sai 1 chỗ:

\(\sqrt{24-3\sqrt{15}}-\sqrt{36-9\sqrt{15}}=\sqrt{\frac{48-6\sqrt{15}}{2}}-\sqrt{\frac{72-18\sqrt{15}}{2}}\)

\(=\sqrt{\frac{\left(\sqrt{3}-3\sqrt{5}\right)^2}{2}}-\sqrt{\frac{\left(3\sqrt{3}-3\sqrt{5}\right)^2}{2}}=\frac{3\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\frac{3\sqrt{5}-3\sqrt{3}}{\sqrt{2}}\)

\(=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

ta có điều kiện của căn thức là : \(\left|4x-9\right|\ge0\)luôn đúng

Vậy điều kiện xác định là \(x\in R\)

\(N=\left(\frac{3}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}-6}{x-2\sqrt{x}}+\frac{1}{x}\right)\)ĐK : \(x>0;x\ne4\)

\(=\left(\frac{3\sqrt{x}-6-\sqrt{x}-2}{x-4}\right):\left(\frac{\sqrt{x}-6}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{x\left(\sqrt{x}-2\right)}\right)\)

\(=\left(\frac{2\sqrt{x}-8}{x-4}\right):\left(\frac{x-6\sqrt{x}-\sqrt{x}+2}{x\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{2x\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\left(x-4\right)\left(x-7\sqrt{x}+2\right)}=\frac{2x\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(x-7\sqrt{x}+2\right)}\)

\(\frac{2x}{x-3}\sqrt{9\left(x-3\right)^2}-2x=\frac{2x}{x-3}\cdot3\left(3-x\right)-2x=-6x-2x=-8x\)(vì x < 0)

\(=\frac{2x}{x-3}\cdot3\left(3-x\right)-2x=\frac{-6x\left(x-3\right)}{x-3}-2x=-6x-2x=-8x\)

mn tìm đkxđ và rút gọn hộ e vs ạ

đề có sai sót khong zạy bạn =))

Với x >= 1 A = x - ( x - 1 ) = x - x + 1 = 1

Với x \(\ge1\)biểu thức A có dạng 

\\(A=x-\left(x-1\right)=x-x+1=1\)

\(1,\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\)

\(\frac{\sqrt{4+2\sqrt{3}}}{2}=\frac{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}{2}\)

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

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

\(2,\sqrt{3-\sqrt{5}}:\sqrt{2}\)

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

\(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{2}\)

\(\frac{\sqrt{5}-1}{2}\)

\(3,\frac{\sqrt{6+2\sqrt{5}}}{1+\sqrt{5}}\)

\(\frac{\sqrt{\sqrt{5}^2+2\sqrt{5}+1}}{1+\sqrt{5}}\)

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

\(4,\sqrt{27+7\sqrt{5}}:\sqrt{2}\)

\(\frac{\sqrt{54+2.7\sqrt{5}}}{2}\)

\(\frac{\sqrt{7^2+2.7.\sqrt{5}+5}}{2}\)

\(\frac{\sqrt{\left(7+\sqrt{5}\right)^2}}{2}\)

\(=\frac{7+\sqrt{5}}{2}\)