Ta có : \(1=\frac{\sqrt{x}+3}{\sqrt{x}+3}\)mà \(\sqrt{x}+5>\sqrt{x}+3\)

Vậy P > 1 

ĐK : x >= 0

Xét hiệu P - 1 ta có : \(P-1=\frac{\sqrt{x}+5}{\sqrt{x}+3}-1=\frac{\sqrt{x}+5-\sqrt{x}-3}{\sqrt{x}+3}=\frac{2}{\sqrt{x}+3}>0\forall x\ge0\)

=> P - 1 > 0 <=> P > 1

34567890 - 234567890   =  - 200000000

Với x = 9 tmdk thay vào A ta được : \(A=\frac{\sqrt{9}+3}{9-4}=\frac{3+3}{5}=\frac{6}{5}\)

\(B=\frac{-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{1}{\sqrt{x}+2}\)

\(P=\frac{A}{B}=\frac{\frac{\sqrt{x}+3}{x-4}}{\frac{1}{\sqrt{x}+2}}=\frac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{1}=\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

Xét hiệu P - 1 ta có : \(P-1=\frac{\sqrt{x}+3}{\sqrt{x}-2}-1=\frac{\sqrt{x}+3-\sqrt{x}+2}{\sqrt{x}-2}=\frac{5}{\sqrt{x}-2}>0\forall x>4\)

=> P > 1

a, Thay x = 9 vào A ta được : \(A=\frac{3+3}{9-4}=\frac{6}{5}\)

b, Với \(x\ge0;x\ne4\)

\(B=-\frac{4}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{-4+\sqrt{x}+2}{x-4}=\frac{\sqrt{x}-2}{x-4}=\frac{1}{\sqrt{x}+2}\)

c, với x > 4 Ta có : \(P=\frac{A}{B}\Rightarrow\frac{\sqrt{x}+3}{x-4}:\frac{1}{\sqrt{x}+2}=\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

Ta có : \(1=\frac{\sqrt{x}-2}{\sqrt{x}-2}\)mà \(\sqrt{x}+3>\sqrt{x}-2\)

Vậy P > 1 

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 

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 \(A=\frac{21}{10}\)

\(\)

\(ĐK:x\ge2;y\le1;z\ge-3\)

\(4x-y+z+10=4\sqrt{x-2}+6\sqrt{1-y}+4\sqrt{z+3}\)

\(\Leftrightarrow4x-y+z+10-4\sqrt{x-2}-6\sqrt{1-y}-4\sqrt{z+3}=0\)

\(\Leftrightarrow\left(4x-8-4\sqrt{x-2}+1\right)+\left(9-6\sqrt{1-y}+1-y\right)+\left(z+3-4\sqrt{z+3}+4\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x-2}-1\right)^2+\left(3-\sqrt{1-y}\right)^2+\left(\sqrt{z+3}-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{x-2}-1=0\\3-\sqrt{1-y}=0\\\sqrt{z+3}-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{9}{4}\\y=-8\\z=1\end{cases}}\left(tm\right)\)

\(\Rightarrow4x+y+z=4\cdot\frac{9}{4}-8+1=2\)

a > 0 thì nó là ĐK luôn rồi bạn

\(A=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1\)

\(=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)