\(S=\sqrt{x-3}+\sqrt{y-4}=1.\sqrt{x-3}+1.\sqrt{y-4}\)

\(\le\sqrt{\left(1^2+1^2\right)\left(x-3+y-4\right)}\)

\(=\sqrt{2}\)

Dấu \(=\)khi \(\hept{\begin{cases}x-3=y-4\\x+y=8\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{7}{2}\\y=\frac{9}{2}\end{cases}}\)

Nếu tìm min 

Ta có :

\(s=\sqrt{x-3}+\sqrt{y-8}\)

 \(\Rightarrow S^2=x-3+4-x+s\sqrt{\left(x-3\right)\left(4-x\right)}\)    

\(=1+\sqrt{\left(x-3\right)\left(4-x\right)}\)\(\ge\)\(1\)

\(\Rightarrow S\ge1\)

 Dấu \(=\)xảy ta \(\Leftrightarrow\hept{\begin{cases}\left(x-3\right)\left(4-x\right)=0\\x+y=8\end{cases}}\) 

\(\sin^2a+\cos^2a=1\)

\(\cos^2a=1-\frac{16}{25}\)

\(\cos^2a=\frac{9}{25}\)

\(\cos a=\frac{3}{5}\)

\(\tan a=\frac{0,8}{\frac{3}{5}}=\frac{4}{3}\)

Trả lời:

\(x-\sqrt{x}-2=8\) \(\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{9}{4}=8\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}=8\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{41}{4}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}-\frac{\sqrt{41}}{2}\right)\left(\sqrt{x}-\frac{1}{2}+\frac{\sqrt{41}}{2}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{1+\sqrt{41}}{2}\right)\left(\sqrt{x}-\frac{1-\sqrt{41}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-\frac{1+\sqrt{41}}{2}=0\\\sqrt{x}-\frac{1-\sqrt{41}}{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\frac{1+\sqrt{41}}{2}\\\sqrt{x}=\frac{1-\sqrt{41}}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\left(\frac{1+\sqrt{41}}{2}\right)^2\\x=\left(\frac{1+\sqrt{41}}{2}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{21+\sqrt{41}}{2}\left(tm\right)\\x=\frac{21-\sqrt{41}}{2}\left(tm\right)\end{cases}}\)

Vậy ...

\(x-\sqrt{x}-2=8\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

phương trình có dạng : \(t^2-t-10=0\)

\(\Delta=1-4\left(-10\right)=41>0\)

pt có 2 nghiệm phân biệt 

\(t_1=\frac{1-\sqrt{41}}{2}\left(ktm\right);t_2=\frac{1+\sqrt{41}}{2}\)

theo cách đặt \(x=\frac{42+2\sqrt{41}}{4}=\frac{21+\sqrt{41}}{2}\)

Bổ sung ĐK : x >= 0 

\(\sqrt{x-4\sqrt{x}+4}=\sqrt{\left(\sqrt{x}-2\right)^2}=\left|\sqrt{x}-2\right|=\sqrt{x}-2\)

sửa Với đk x >= 0 

thì \(\left|x-2\right|=\left|2-x\right|=2-x\)

\(y=\left(2m-5\right)x+m-4\)(d) ; \(y=3x+7\)(d1)

Thay x = 2 vào ptđt (d1) ta được : 

\(y=6+7=13\)

Vậy (d) cắt (d1) tại A(2;13) 

(d) cắt (d1) tại A(2;13) <=> \(2\left(2m-5\right)+m-4=13\Leftrightarrow5m-14=13\Leftrightarrow m=\frac{27}{5}\)

Ta có : \(sin^2\alpha=1-cos^2\alpha\)

\(=1-\left(\frac{15}{17}\right)^2\)

\(=\frac{64}{289}\)

\(\Rightarrow sin\alpha=\sqrt{\frac{64}{289}}=\frac{8}{17}\)(do \(sin\alpha>0\))

Ta lại có : \(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{\frac{8}{17}}{\frac{15}{17}}=\frac{8}{15}\)