a) \(\sqrt{2x+\sqrt{6x^2+1}}=x+1\)

\(\Leftrightarrow2x+\sqrt{6x^2+1}=x^2+2x+1\)

\(\Leftrightarrow\sqrt{6x^2+1}=x^2+1\)

\(\Leftrightarrow6x^2+1=x^4+2x^2+1\)

\(\Leftrightarrow x^4-4x^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

\(B=x^5-x+7\)

\(B=x\left(x^4-1\right)+6+1\)

\(B=x\left(x^4-x^2+x^2-1\right)+6+1\)

\(B=x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)+6+1\)

Ta có: \(x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)+6\)chia hết cho 3

=> B chia 3 dư 1

=> B không phải là scp với mọi x thuộc Z⁺( đpcm )

\(B=\sqrt{7-2\sqrt{12}}\)

\(B=\sqrt{4-2.2.\sqrt{3}+3}\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(B=\left|2-\sqrt{3}\right|\)

\(B=2-\sqrt{3}\)

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

N = \(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

N = \(\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

N = \(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

N = \(\frac{\sqrt{x}}{\sqrt{x}-2}\)

b) Với x \(\ge\)0; x \(\ne\)4

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

=> \(-3\sqrt{x}=\sqrt{x}-2\)

<=> \(-4\sqrt{x}=-2\)

<=> \(\sqrt{x}=\frac{1}{2}\)

<=> \(x=\frac{1}{4}\)

c) x = 25 => N = \(\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{5-3}=\frac{5}{2}\)

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

\(N=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(N=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(N=\frac{\left(\sqrt{x}+2\right)\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(N=\frac{\sqrt{x}}{\sqrt{x}-2}\)

b) \(N=-\frac{1}{3}\)

\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=-\frac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=2\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)

c) \(N=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{5-2}=\frac{5}{3}\)

\(P=\frac{\sqrt{x}-1}{\sqrt{x}+9}=\frac{\sqrt{x}+9-10}{\sqrt{x}+9}=1-\frac{10}{\sqrt{x}+9}\)

Để \(\sqrt{p}< \frac{1}{3}\)thì\(P< \frac{1}{9}\)hay\(1-\frac{10}{\sqrt{x}+9}< \frac{1}{9}\Leftrightarrow\frac{8}{9}< \frac{10}{\sqrt{x}+9}\Leftrightarrow\frac{10}{11,25}< \frac{10}{\sqrt{x}+9}\Leftrightarrow\sqrt{x}+9>11,25\)

\(\Leftrightarrow\sqrt{x}>2,25\Leftrightarrow x>\frac{81}{16}\)

\(A=\frac{\left(x-9\right)+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)\(=\left(\sqrt{x}+3\right)+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{25}{\sqrt{x}+3}}-6=2.5-4=6\)

Dấu'=' xảy ra khi và chỉ khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\)

\(\Rightarrow\left(\sqrt{x}+3\right)^2=25\Rightarrow\sqrt{x}+3=5\left(do\sqrt{x}+3>0\right)\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

Vậy MinA=4 khi và chỉ khi x=4

+) \(2^{2n}=4^n=4\left(4^{n-1}-1\right)+4\)với \(n\inℕ^∗\)

+) \(4^{n-1}-1\equiv1-1\equiv0\)(mod 3)

\(\Rightarrow4\left(4^{n-1}-1\right)⋮12\)

Vậy \(2^{2n}\)chia 12 dư 4