\(a,ĐK:x\ge-5\\ A=2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=3\sqrt{x+5}\\ b,A=6\Leftrightarrow\sqrt{x+5}=\dfrac{6}{3}=2\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)

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

\(A=3\sqrt{x+5}\)

\(3\sqrt{x+5}=6\)

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

\(\left\{{}\begin{matrix}2\ge0\left(ld\right)\\x+5=4\end{matrix}\right.\)

\(x=-1\)

a: \(A=\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}\)

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

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

b: A=6

=>\(3\sqrt{x+5}=6\)

=>\(\sqrt{x+5}=2\)

=>x+5=4

=>x=-1

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

Câu 3: 

=>|2x-1|=x+10

TH1: x>=1/2

=>2x-1=x+10

=>x=11(nhận)

TH2: x<1/2

=>1-2x=x+10

=>-3x=9

=>x=-3(nhận)

\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)