\(ĐK:\frac{-1}{4}\le x\le3\)

\(6\sqrt{4x+1}+2\sqrt{3-x}=3x+14\)\(\Leftrightarrow\left(4x+1-6\sqrt{4x+1}+9\right)+\left(3-x-2\sqrt{3-x}+1\right)=0\)\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2=0\)(*)

Xảy ra (*)\(\Leftrightarrow\hept{\begin{cases}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{cases}}\Leftrightarrow x=2\left(t/m\right)\)

Vậy nghiệm duy nhất của phương trình là 2.

\(A=\left(\frac{\sqrt{x}}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}-1}{x-1}\)

\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{x-1}\)

\(=\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1}{\sqrt{x}-1}\)

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

CM: \(a=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{\sqrt{2}}{8}\Rightarrow a+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)

\(\Leftrightarrow\left(a+\frac{\sqrt{2}}{8}\right)^2=\left(\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\right)^2\)\(\Leftrightarrow a^2+\frac{a\sqrt{2}}{4}+\frac{1}{32}=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\Leftrightarrow a^2+\frac{2\sqrt{a}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)

\(\Leftrightarrow4a^2+\sqrt{2}a-\sqrt{2}=0\)

Theo trên: \(4a^2+\sqrt{2}a-\sqrt{2}=0\Rightarrow a^2=\frac{\sqrt{2}\left(1-a\right)}{4}\Rightarrow a^4=\frac{a^2-2a+1}{8}\)

\(\Rightarrow a^4+a+1=\frac{a^2-2a+1}{8}+a+1=\left(\frac{a+3}{2\sqrt{2}}\right)^2\)

\(B=a^2+\sqrt{a^4+a+1}=a^2+\frac{a+3}{2\sqrt{2}}=\frac{2\sqrt{2}a^2+a+3}{2\sqrt{2}}\)\(=\frac{4a^2+\sqrt{2}a+3\sqrt{2}}{4}=\frac{4\sqrt{2}}{4}=\sqrt{2}\)

Ta có : S = 4 + 4² + 4³ + 4⁴ + 4⁵ + 4⁶ + ... + 4²⁰¹⁴ + 4²⁰¹⁵ + 4²⁰¹⁶

= (4 + 4² + 4³) + (4⁴ + 4⁵ + 4⁶) + ... + (4²⁰¹⁴ + 4²⁰¹⁵ + 4²⁰¹⁶)

= 4(1 + 4 + 4²) + 4⁴(1 + 4 + 4²) + ... + 4²⁰¹⁵(1 + 4 + 4²)

= (1 + 4 + 4²)(4 + 4⁴ + ... + 4²⁰¹⁵)

= 21(4 + 4⁴ + ... + 4²⁰¹⁵) 

=> S \(⋮\)21 (1)

Lại có S = 4 + 4² + 4³ + 4⁴ + .... + 4²⁰¹⁵ + 4²⁰¹⁶

= (4 + 4²) + (4³ + 4⁴) + .... + (4²⁰¹⁵ + 4²⁰¹⁶)

= (4 + 4²) + 4²(4 + 4²) + ... + 4²⁰¹⁴(4 + 4²)

= (4 + 4²)(1 + 4² + ... + 4²⁰¹⁴)                         

= 20(1 + 4² + ... + 4²⁰¹⁴)  

=> S \(⋮\)20 (2)

Lại có ƯCLN(20;21) = 1 (3)

Từ (1)(2)(3)

=> S \(⋮20.21\Rightarrow S⋮420\)(ĐPCM)

24:28ĐANG PHÁT

Đk:  0 < x;y;z < = 1

Ta có:

\(x\sqrt{1-y^2}+y\sqrt{1-z^2}+z\sqrt{1-x^2}=\frac{3}{2}\)

<=> \(2x\sqrt{1-y^2}+2y\sqrt{1-z^2}+2z\sqrt{1-x^2}=3\)

<=> \(3-2x\sqrt{1-y^2}-2y\sqrt{1-z^2}-2z\sqrt{1-x^2}=0\)

<=> \(1-y^2-2x\sqrt{1-y^2}+x^2+1-z^2-2y\sqrt{1-z^2}+y^2+1-x^2-2z\sqrt{1-x^2}+z^2=0\)

<=> \(\left(\sqrt{1-y^2}-x\right)^2+\left(\sqrt{1-z^2}-y\right)^2+\left(\sqrt{1-x^2}-z\right)^2=0\)

<=> \(\hept{\begin{cases}\sqrt{1-y^2}-x=0\\\sqrt{1-z^2}-y=0\\\sqrt{1-x^2}-z=0\end{cases}}\) <=> \(\hept{\begin{cases}\sqrt{1-y^2}=x\\\sqrt{1-z^2}=y\\\sqrt{1-x^2}=z\end{cases}}\) <=> \(\hept{\begin{cases}1-y^2=x^2\left(1\right)\\1-z^2=y^2\left(2\right)\\1-x^2=z^2\left(3\right)\end{cases}}\)

Từ (1), (2) và (3) cộng vế theo vế:

\(3-\left(x^2+y^2+z^2\right)=x^2+y^2+z^2\) <=> \(2\left(x^2+y^2+z^2\right)=3\) <=> \(x^2+y^2+z^2=\frac{3}{2}\)