- Đề sai nhiều vậy sửa lại đi bạn ;-;

e) Ta có: \(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{2}{3}\right\}\)

\(a,\dfrac{3}{\sqrt{12x-1}}\) xác định \(\Leftrightarrow12x-1>0\Leftrightarrow12x>1\Leftrightarrow x>\dfrac{1}{12}\)

\(b,\sqrt{\left(3x+2\right)\left(x-1\right)}\) xác định \(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}3x+2\ge0\\x-1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}3x+2\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\)

\(c,\sqrt{3x-2}.\sqrt{x-1}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3x-2\ge0\\x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)

\(d,\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) xác định \(\Leftrightarrow-x+5>0\Leftrightarrow x< 5\)

a: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)

\(=\left(3-3^2+3^3-3^4\right)+\left(3^5-3^6+3^7-3^8\right)+\cdots+\left(3^{21}-3^{22}+3^{23}-3^{24}\right)\)

\(=\left(3-3^2+3^3-3^4\right)+3^4\left(3-3^2+3^3-3^4\right)+\cdots+3^{20}\left(3-3^2+3^3-3^4\right)\)

\(=-60\left(1+3^4+\cdots+3^{20}\right)\) ⋮60

\(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)

\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+\cdots-\left(3^{22}-3^{23}+3^{24}\right)\)

\(=\left(3-3^2+3^3\right)-3^3\left(3-3^2+3^3\right)+\cdots-3^{21}\left(3-3^2+3^3\right)\)

\(=21\left(1-3^3+\cdots-3^{21}\right)\) ⋮21

=>C⋮7

mà C⋮60

và ƯCLN(60;7)=1

nên C⋮60*7

=>C⋮420

b:

ĐKXĐ: y>=1

\(\left(x+1\right)^{2022}\ge0\forall x;\left(\sqrt{y-1}\right)^{2023}\ge0\forall y\) thỏa mãn ĐKXĐ

=>\(\left(x+1\right)^{2022}+\left(\sqrt{y-1}\right)^{2023}\ge0\forall x,y\)

Dấu '=' xảy ra khi x+1=0 và y-1=0

=>x=-1 và y=1

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

24: \(-6x+5\sqrt{x}+1=\left(\sqrt{x}-1\right)\left(-6\sqrt{x}-1\right)\)

21: \(x^2-3x\sqrt{y}+2y\)

\(=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)

\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)

\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)

23: \(\sqrt{x^3}-2\sqrt{x}-x\)

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

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

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

Lời giải:

ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}-\sqrt{(x-1)-2\sqrt{x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}-\sqrt{(\sqrt{x-1}-1)^2}=2$

$\Leftrightarrow |\sqrt{x-1}+1|-|\sqrt{x-1}-1|=2$
Nếu $2\geq x\geq 1$ thì:

$\sqrt{x-1}+1+(1-\sqrt{x-1})=2$
$\Leftrightarrow 2=2$ (luôn đúng)

Nếu $x>2$ thì: $\sqrt{x-1}+1+(\sqrt{x-1}-1)=2$
$\Leftrightarrow 2\sqrt{x-1}=2$
$\Leftrightarrow x-1=1$

$\Leftrihgtarrow x=2$ (loại)

Vậy $2\geq x\geq 1$

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