\(a,9234:\left[3\cdot3\cdot\left(1+8^3\right)\right]\\ =9234:\left[9\cdot\left(1+512\right)\right]\\ =9234:\left(9\cdot513\right)\\ =9234:4617\\ =2\\ b,76-\left\{2\cdot\left[2\cdot5^2-\left(31-2\cdot3\right)\right]\right\}+3\cdot25\\ =76-\left\{2\cdot\left[2\cdot25-\left(31-6\right)\right]\right\}+75\\ =76-\left[2\cdot\left(50-25\right)\right]+75\\ =76-\left(2\cdot25\right)+75\\ =76-50+75\\ =26+75\\ =-49\)

\(2^{x+1}-2^x=32\\ \Rightarrow2^x\cdot2^1-2^x=32\\ \Rightarrow2^x\left(2-1\right)=32\\ \Rightarrow2^x=32:1\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)

\(\left|x\right|-1\dfrac{2}{3}=0\\ \Rightarrow\left|x\right|=0+\dfrac{5}{3}\\ \Rightarrow\left|x\right|=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Mà `x>0`

`=> x=5/3`

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\(2\left|x\right|-3=11\\ \Rightarrow2\left|x\right|=11+3\\ \Rightarrow2\left|x\right|=14\\ \Rightarrow\left|x\right|=\dfrac{14}{2}=7\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

\(45^{10}\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}\\ =3^{20}\cdot5^{10}\cdot5^{30}\\ =3^{20}\cdot5^{40}\)

em làm mà onl ít qua, onl nhiều lên^^

\(\dfrac{45}{15}-\left\{\dfrac{1}{2}+\left[\dfrac{1}{3}+\left(\dfrac{1}{4}\right)^{-1}\right]^{-1}\right\}^{-1}\)

\(=3-\left[\dfrac{1}{2}+\left(\dfrac{1}{3}+4\right)^{-1}\right]^{-1}\\ =3-\left(\dfrac{1}{2}+\dfrac{3}{13}\right)^{-1}\\ =3-\left(\dfrac{19}{26}\right)^{-1}\\ =3-\dfrac{26}{19}\\ =\dfrac{31}{19}\)

\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

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

Ta có :

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

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C làm giống câu a, nhé.

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

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\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)

\(\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4\)

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\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}\)

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\(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}\)