Đề bài là rút gọn A hả bạn?  Đkxđ : a > 0 
\(A=\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)\(.\left(\frac{1+\sqrt{a}}{1-a}\right)^2\) \(=\frac{1+a\sqrt{a}-\sqrt{a}\left(1+\sqrt{a}\right)}{1+\sqrt{a}}\)\(.\left(\frac{1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\) \(=\frac{1+a\sqrt{a}-\sqrt{a}-a}{1+\sqrt{a}}\)\(.\left(\frac{1}{1-\sqrt{a}}\right)^2\)\(=\frac{\sqrt{a}\left(1-a\right)+\left(1-a\right)}{1+\sqrt{a}}\)\(.\left(\frac{1}{1-\sqrt{a}}\right)^2\)\(=\frac{\left(\sqrt{a}+1\right)\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1+\sqrt{a}}\)\(.\left(\frac{1}{1-\sqrt{a}}\right)^2\)=  \(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)1^2}{\left(1-\sqrt{a}\right)^2}\)
= \(\frac{1+\sqrt{a}}{1-\sqrt{a}}\)

\(\sqrt{x-3}+\sqrt{5-x}=x^2-8x+18.\)

ĐK: \(3\le x\le5\)

\(PT\Leftrightarrow\sqrt{x-3}-1+\sqrt{5-x}-1=x^2-8x+18-2\) 

       \(\Leftrightarrow\frac{x-3-1}{\sqrt{x-3}-1}+\frac{5-x-1}{\sqrt{5-x}+1}=\left(x-4\right)^2\)

      \(\Leftrightarrow\frac{x-4}{\sqrt{x-3}+1}+\frac{4-x}{\sqrt{5-x}+1}=\left(x-4\right)^2\)

       \(\Leftrightarrow\left(x-4\right)^2-\frac{x-4}{\sqrt{x-3}+1}+\frac{x-4}{\sqrt{5-x}+1}=0\)

        \(\Leftrightarrow\left(x-4\right).\left(x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}\right)=0\)

         \(\Rightarrow\orbr{\begin{cases}x-4=0\\x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}=0\end{cases}}\)

           \(\Rightarrow\orbr{\begin{cases}x=4\left(TM\right)\\x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}=0\end{cases}}\)  (Vô nghiệm)

Vậy pt có nghiệm x-4

cảm ơn huệ lam

2020.2019^5 = (2019+1).2019^5 = 2019^6+2019^5 làm tương tự với các x còn lại

A= 2019^6 - 2019^6 +.....-2019^2-2019 +2020 = 1 vậy A=1

ta có x = 2019 \(\Rightarrow\)2020 = x+1  

thay 2020 = x+1 vào A ta có

\(A=x^6-\left(x+1\right).x^5+\left(x+1\right).x^4-...-\left(x+1\right).x+2020\)

\(=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2020\)

\(=-x+2020\)

\(=-2019+2020\)

\(=1\)

vậy A = 1

học tốt !!!

Bình phương 2 vế lên bn

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

=> \(\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5-6\sqrt{2x-5}+9}=6\)

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

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

 Đến đây dễ bạn rồi tự làm tiếp

a)\(-\frac{2}{\sqrt{1-3x}}\text{có nghĩa }\Leftrightarrow1-3x>0\)

\(\Leftrightarrow-3x>-1\Leftrightarrow x< 1\)

b)\(\sqrt{\frac{-5}{x^2+6}}\text{có nghĩa }\Leftrightarrow\frac{-5}{x^2+6}\ge0;x^2+6\ne0\)

\(\Leftrightarrow x^2+6< 0\Leftrightarrow x^2< -6\left(\text{vô lí }\right)\)

\(x\in\varnothing\)

\(\sqrt{x+5}+\frac{1}{x+5}\text{có nghĩa }\Leftrightarrow x+5>0\)

\(\Leftrightarrow x>-5\)

\(\sqrt{\left(x-1\right)\left(x-2\right)}\text{có nghĩa }\Leftrightarrow\left(x-1\right)\left(x-2\right)\ge0\)

TH1: \(\left(x-1\right)\ge0\text{ và }\left(x-2\right)\ge0\)

\(\Rightarrow x\ge2\)

TH2: \(\left(x-1\right)\le0\text{ và }\left(x-2\right)\le0\)

\(\Rightarrow x\le1\)

\(\sqrt{2x-3}\text{có nghĩa }\Leftrightarrow2x-3\ge0\)

\(\Leftrightarrow2x\ge3\Leftrightarrow x\ge\frac{3}{2}\)

\(\sqrt{1-2x}\text{có nghĩa }\Leftrightarrow1-2x\ge0\)

\(\Leftrightarrow1\ge2x\Leftrightarrow x\le\frac{1}{2}\)

\(\sqrt{x^2}\text{có nghĩa }\Leftrightarrow x^2\ge0\text{luôn đúng }\forall x\)

\(\sqrt{\frac{-1}{1-x}}\text{có nghĩa }\frac{-1}{1-x}\ge0;x\ne1\)

\(\Leftrightarrow1-x< 0\Leftrightarrow x>1\)

khi nào luôn đúng ạ