Đặt A = \(\sqrt{\frac{3+\sqrt{5}}{\sqrt{3}+\sqrt{5}}}+\sqrt{\frac{3-\sqrt{5}}{\sqrt{3}+\sqrt{5}}}\)

A = \(\frac{\sqrt{3+\sqrt{5}}}{\sqrt{\sqrt{3}+\sqrt{5}}}+\frac{\sqrt{3-\sqrt{5}}}{\sqrt{\sqrt{3}+\sqrt{5}}}\)

A = \(\frac{\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{\sqrt{3}+\sqrt{5}}}\)

A² = \(\frac{\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)^2}{\left(\sqrt{\sqrt{3}+\sqrt{5}}\right)^2}\)

A² = \(\frac{3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}{\sqrt{3}+\sqrt{5}}\)

A² = \(\frac{6+2\sqrt{6-5}}{\sqrt{3}+\sqrt{5}}\)

A² = \(\frac{8\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}\)

A² = \(\frac{8\left(\sqrt{3}-\sqrt{5}\right)}{3-5}=-4\left(\sqrt{3}-\sqrt{5}\right)\)

A² = \(4\left(\sqrt{5}-\sqrt{3}\right)\)

=> A = \(2.\sqrt{\sqrt{5}-\sqrt{3}}\)

P/s : làm bừa thôi!

\(\sqrt{x-2018}+\sqrt{x^2+11}+x^2=\sqrt{y^2+11}+\sqrt{y-2018}+y^2\)

\(\Leftrightarrow x=y\)

\(\Rightarrow M=x^{11}-x^{2018}\)

Đến đây em tịt !!

ĐK : \(x\ge-2\)

PT <=> \(x^2=4\left(x+2\right)\)

\(x^2=4x+8\)

\(x^2-4x-8=0\)

\(\Delta=\left(-4\right)^2-4.\left(-8\right)=16+32=48>0\)

\(x_1=\frac{4-\sqrt{48}}{2}\left(ktm\right);x_2=\frac{4+\sqrt{48}}{2}\left(tm\right)\)

x² = 4(x+2)

x²=4x+8

x²-4x-8=0

x²-4x+4-12=0

(x-2)²=12

x-2=\(\sqrt{12}\)or x-2=\(-\sqrt{12}\)

Xong bạn tính x nha

Ta có: \(A=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right).\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)    (   ĐK: \(x\ne0,\)\(x\ne9,\)\(x\ge3\))

     \(\Leftrightarrow A=\frac{\sqrt{x}.\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3\sqrt{x}-9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)

     \(\Leftrightarrow A=\frac{3.\left(2\sqrt{x}+4\right)}{\left(9-x\right).\sqrt{x}}\)

     \(\Leftrightarrow A=\frac{6\sqrt{x}+12}{9\sqrt{x}-x}\)

:V

Câu đầu cho x > 0 thì dễ hơn ...... 

Sử dụng BĐT AM - GM ta dễ có:\(D=\sqrt{x}+\frac{9}{\sqrt{x}+2}=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-2\ge2\sqrt{\left(\sqrt{x}+2\right)\cdot\frac{9}{\sqrt{x}+2}}-2=4\)

Đẳng thức xảy ra tại x=1

\(E=\frac{x+1}{\sqrt{x}}\ge\frac{2\sqrt{x}}{\sqrt{x}}=2\) Đẳng thức xảy ra tại x=1

Làm 2 cái thôi còn lại tương tự bạn nhé :) 

+ Ta có: \(D=\sqrt{x}+\frac{9}{\sqrt{x}+2}\)

       \(D=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-2\)

   Áp dụng bất đẳng thức Cô-si cho phương trình \(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\) ta có: 

         \(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge\sqrt{\left(\sqrt{x}+2\right).\left(\frac{9}{\sqrt{x}+2}\right)}=\sqrt{9}=3\)

         \(\Rightarrow\)\(D\ge3-2=1\)

   Dấu bằng xảy ra khi và chỉ khi: \(\sqrt{x+2}=\frac{9}{\sqrt{x}+2}\)

                                               \(\Leftrightarrow\left(\sqrt{x}+2\right)^2=9\)

                                               \(\Leftrightarrow\sqrt{x}+2=\pm3\)

                                               \(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+2=-3\\\sqrt{x}+2=3\end{cases}}\)

                                               \(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-5\left(L\right)\\\sqrt{x}=1\end{cases}}\)

                                               \(\Leftrightarrow x=\pm1\)

 Vậy \(S=\left\{\pm1\right\}\)

a, Ta có : \(\sqrt{120}^2=120\)

\(\left(5\sqrt{7}\right)^2=25.7=175\)

\(\Rightarrow\sqrt{120}< 5\sqrt{7}\)

b, Ta có : \(\left(\frac{1}{6}\sqrt{5}\right)^2=\frac{1}{36}.5=\frac{5}{36}\)

\(\left(\frac{1}{5}\sqrt{6}\right)^2=\frac{1}{25}.6=\frac{6}{25}\)

\(\Rightarrow\frac{5}{36}< \frac{6}{25}\)

a) \(\sqrt{60}-\sqrt{135}+\frac{1}{3}\sqrt{15}\)

\(=2\sqrt{15}-3\sqrt{15}+\frac{1}{3}\sqrt{15}\)

\(=-\frac{2}{3}\sqrt{15}\)

b) \(\sqrt{28}-\frac{1}{2}\sqrt{343}+2\sqrt{63}\)

\(=2\sqrt{7}-\frac{7}{2}\sqrt{7}+6\sqrt{7}\)

\(=\frac{9}{2}\sqrt{7}\)

c) \(\sqrt{12}-\frac{2}{3}\sqrt{27}+\sqrt{243}\)

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

\(=9\sqrt{3}\)

\(\sin\alpha=\frac{8}{17}\Rightarrow sin^2\alpha=\frac{64}{289}\Rightarrow cos^2\alpha=1-sin^2\alpha=1-\frac{64}{289}=\frac{225}{289}\)

\(\Rightarrow cos\alpha=\frac{15}{17}\)

từ đó tính ra \(tan\alpha;cot\alpha\)

Ta có: \(\sin^2\alpha+\tan^2\alpha=1\)

\(\Leftrightarrow\frac{64}{289}+\tan^2\alpha=1\)

\(\Leftrightarrow\tan^2\alpha=\frac{225}{289}\)

\(\Rightarrow\tan\alpha=\frac{15}{17}\)

Đến đây thì dễ rồi:

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{15}{8}\) ; \(\cot\alpha=\frac{8}{15}\)

Hình vẽ chung cho cả ba bài.

Bài 1:

\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{15^2}+\frac{1}{20^2}=\frac{1}{144}\)

\(\Rightarrow AH^2=144\Rightarrow AH=12\)

\(BH=\sqrt{AB^2-AH^2}=\sqrt{15^2-12^2}=\sqrt{81}=9\)

\(CH=\sqrt{AC^2-AH^2}=\sqrt{20^2-12^2}=\sqrt{256}=16\)

\(\Rightarrow BC=BH+CH=9+16=25\)

Bài 2,3 bạn nhìn hình vẽ và sử dụng hệ thức lượng để tính tiếp như bài 1.

Bài 2:                                                    Bài giải

Đặt BH = x (0 < x < 25) (cm) => CH = 25 - x (cm)

Ta có : \(AH^2=BH\cdot CH\text{ }\Rightarrow\text{ }x\left(25-x\right)=144\text{ }\Rightarrow\text{ }x^2-25x+144=0\)

\(\left(x-9\right)\left(x-16\right)=0\text{ }\Rightarrow\orbr{\begin{cases}x=9\\x=16\end{cases}}\left(tm\right)\)

Nếu BH = 9 cm thì CH = 16 cm \(\Rightarrow\text{ }AB=\sqrt{AH^2+BH^2}=\sqrt{9^2+12^2}=15\text{ }\left(cm\right)\)

\(AC=\sqrt{AH^2+CH^2}=\sqrt{12^2+16^2}=20\text{ }\left(cm\right)\)

Nếu BH = 16 cm thì CH = 9 cm

\(\Rightarrow\text{ }AB=\sqrt{AH^2+BH^2}=\sqrt{12^2+16^2}=20\text{ }\left(cm\right)\)

\(AC=\sqrt{AH^2+CH^2}=\sqrt{9^2+12^2}=15\text{ }\left(cm\right)\)