\(x+5-5\sqrt{x-1}=0\)

\(\Leftrightarrow x+5=5\sqrt{x-1}\)

Điều kiện :

\(\hept{\begin{cases}x+5\ge0\\x-1\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-5\\x\ge1\end{cases}}\Leftrightarrow x\ge1}\)

Bình phương 2 vế ta có :

\(\left(x+5\right)^2=25\left(x-1\right)\)

\(\Leftrightarrow x^2+10x+25=25x-25\)

\(\Leftrightarrow x^2+10x-25x=-25-25\)

\(\Leftrightarrow x^2-15x+50=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=5\end{cases}}\)( thỏa mãn )

Vậy ................

Mình làm hơi tắt bước cuối nên bạn thông cảm nha

Cos A sao > 1 đc bạn ơi . Bn xem lại đề ạ

Tl

Đề bài nỗi ròi

#Kirito

Sorry I can't help im only grade 4

\(\frac{5\sqrt{x}-5}{x-1}=\frac{5}{\sqrt{x}+1}\Rightarrow\sqrt{x}+1\inƯ\left(1;5\right)\)ĐK : \(x\ne1\)

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}}}\)

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

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

\(=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

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

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

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

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

\(sin57^0=cos\left(90^0-57^0\right)=cos33^0\)

\(cos66^0=cos\left(90^0-66^0\right)=cos24^0\)

\(tan77^0=cot\left(90^0-77^0\right)=cot13^0\)

\(cot57^0=tan\left(90^0-57^0\right)=tan33^0\)