1,  \(\sqrt{4x-20}-\sqrt{x+5}=\sqrt{x-5}\)ĐK : x >= 5

\(\Leftrightarrow2\sqrt{x-5}-\sqrt{x+5}-\sqrt{x-5}=0\)

\(\Leftrightarrow\sqrt{x-5}=\sqrt{x+5}\Leftrightarrow x-5=x+5\Leftrightarrow0=10\)( vô lí )

Vậy pt vô nghiệm 

A B C ∝

Do  \(tan.\alpha=3\Rightarrow\frac{AB}{AC}=\frac{1}{3}\)

\(\Rightarrow\frac{AB^2}{AC^2}=\frac{1}{9}\Rightarrow AB^2=\frac{AC^2}{9}=\frac{AB^2+AC^2}{1+9}=\frac{BC^2}{10}\)

\(\Rightarrow AB^2=\frac{AC^2}{9}=\frac{BC^2}{10}\Rightarrow AB=\frac{AC}{3}=\frac{BC}{\sqrt{10}}\)

Đặt \(AB=\frac{AC}{3}=\frac{BC}{\sqrt{10}}=k\)

\(\Rightarrow AB=k;AC=3k;BC=\sqrt{10}k\)

\(\Rightarrow A=sin^2a+2sina.cosa-5cos^2a\)

\(=\left(\frac{AC}{BC}\right)^2+2\frac{AC}{BC}.\frac{AB}{BC}-5\left(\frac{AB}{BC}\right)^2\)

\(=\left(\frac{3k}{\sqrt{10}k}\right)^2+2.\frac{3k}{\sqrt{10}k}.\frac{k}{\sqrt{10}k}-5.\left(\frac{k}{\sqrt{10}k}\right)^2\)

\(=\frac{9}{10}+\frac{2.3}{10}-\frac{5.1}{10}=\frac{21}{10}\)

Vậy \(A=\frac{21}{10}\)

\(\)

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chuyện đời là thế .

A 70 o 120 o B D C d' d 1 3 2 4 1 1

Vì \(\widehat{B_1}=\widehat{B_4}\left(\text{2 góc đối đỉnh}\right)\)\(\Rightarrow\widehat{B_4}=120^o\)

Vì \(\widehat{B_4}=\widehat{C_2}\left(\text{ 2 góc so le trong }\right)\)\(\Rightarrow\widehat{C_2}=120^o\)

Vì \(\widehat{C_2}+\widehat{C_3}=180^o\)( 2 góc kề bù ) \(\Rightarrow120^o+\widehat{C_3}=180^o\)\(\Rightarrow\widehat{C_3}=180^o-120^o=60^o\)

Vì \(\widehat{A_1}=\widehat{D_1}\left(\text{ 2 góc so le trong }\right)\)\(\Rightarrow\widehat{D_1}=70^o\)