\(a,P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}+\dfrac{6\sqrt{x}-4}{1-x}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

Tại \(x=7-4\sqrt{3}\)  Ta có :

 \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{7-4\sqrt{3}}-1}{\sqrt{7-4\sqrt{3}}+1}\\ =\dfrac{\sqrt{4-4\sqrt{3}+3}-1}{\sqrt{4-3\sqrt{3}+3}+1}\\ =\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}-1}{\sqrt{\left(2-\sqrt{3}\right)^2}+1}\\ =\dfrac{2-\sqrt{3}-1}{2-\sqrt{3}+1}\\ =\dfrac{1-\sqrt{3}}{3-\sqrt{3}}\\ =\dfrac{1-\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-1\right)}\\ =\dfrac{-1}{\sqrt{3}}\)

`A=x(3x+12)-(7x-20)+x^2(2x-3)-x(2x^2+5)`

`= 3x^2 + 12x -7x+20 +2x^3 -3x^2 - 2x^3 -5x`

`=(2x^3-2x^3)+(3x^2-3x^2) +(12x-7x-5x) +20`

`=0+0+0+20`

`=20`

Bài `1`

\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\\ =\sqrt{3-2\sqrt{15}+5}-\sqrt{3+2\sqrt{15}+5}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\\ =\left|\sqrt{3}-\sqrt{5}\right|-\left|\sqrt{3}+\sqrt{5}\right|\\ =\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\\ =-2\sqrt{3}\)

\(B=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\\ B-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot2+1^2}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)

Mà \(A-\sqrt{2}=-2\)

\(\Rightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)

\(A=\sqrt{243}-\sqrt{27}+\sqrt{3}-\sqrt{48}\\ =\sqrt{81\cdot3}-\sqrt{9\cdot3}+\sqrt{3}-\sqrt{16\cdot3}\\ =9\sqrt{3}-3\sqrt{3}+\sqrt{3}-4\sqrt{4}\\ =\left(9-3+1-4\right)\sqrt{3}\\ =3\sqrt{3}\)

\(B=\dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{3}+\sqrt{5}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\sqrt{5}-\sqrt{3}\\ =\sqrt{5}+1+\sqrt{3}-\sqrt{5}-\sqrt{3}\\ =1\)

\(\left(-\dfrac{3}{5}\right)^2-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\\ \Rightarrow\dfrac{9}{25}-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\\ \Rightarrow x-\dfrac{1}{3}=\dfrac{9}{25}-\dfrac{4}{25}\\ \Rightarrow x-\dfrac{1}{3}=\dfrac{5}{25}\\ \Rightarrow x=\dfrac{1}{5}+\dfrac{1}{3}\\ \Rightarrow x=\dfrac{3}{15}+\dfrac{5}{15}\\ \Rightarrow x=\dfrac{8}{15}\)

Vậy `x=8/15`

\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ =\dfrac{x-1}{\sqrt{x}}\)

Ta có :

\(12km^2=120000m^2\)

\(4dam^2=400m^2\)

\(12km^24dam^2=120000+400=120400m^2\)

\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\\ \Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)