a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b) \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

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

c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)

\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)

e) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}=2+\dfrac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x\ge0\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)

\(A=\dfrac{1}{2\left(\sqrt{a}+1\right)}-\dfrac{1}{2\left(\sqrt{a}-1\right)}+\dfrac{a^2+1}{a^2-1}\)

\(=\dfrac{\sqrt{a}-1-\sqrt{a}-1}{2\left(a-1\right)}+\dfrac{a^2+1}{a^2-1}\)

\(=\dfrac{-1}{a-1}+\dfrac{a^2+1}{a^2-1}\)

\(=\dfrac{-a-1+a^2+1}{\left(a-1\right)\left(a+1\right)}=\dfrac{a^2-a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{a+1}\)

b: Để A-1/3<0 thì \(\dfrac{a}{a+1}-\dfrac{1}{3}< 0\)

=>3a-a-1<0

=>2a-1<0

hay 0<a<1/2

ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}\\x\le-\sqrt{5}\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x< -12\end{matrix}\right.\)

hình như đề bài bị sai số thì phải bạn ạ

mình giải cứ bị lệch số ấy

Đề khá hay đấy! Nhưng lần sau đừng viết sai đề nx!

a) ĐK: \(x>4\)

b) \(P=\dfrac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{1-\dfrac{8}{x}+\dfrac{16}{x^2}}}\)

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

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

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

= \(\dfrac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\dfrac{4}{x}}\) = \(\left[{}\begin{matrix}\dfrac{2x\sqrt{x-4}}{x-4}khix\ge8\\\dfrac{4x}{x-4}khi4< x< 8\end{matrix}\right.\)

Xét \(P=\dfrac{2x}{\sqrt{x-4}}\left(x\ge8\right)\) thì:

Để \(P\in Z\) khi \(\dfrac{2x-8+8}{\sqrt{x-4}}\in Z\)

<=> \(2.\left(\sqrt{x-4}\right)+\dfrac{8}{\sqrt{x-4}}\in Z\)

<=> \(\left\{{}\begin{matrix}\sqrt{x-4}\in Z^+\\\sqrt{x-4}\inƯ\left(8\right)\end{matrix}\right.\)

Mà \(x\ge8\) => \(\left[{}\begin{matrix}\sqrt{x-4}=2\\\sqrt{x-4}=4\\\sqrt{x-4}=8\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=8\\x=20\\x=68\end{matrix}\right.\)

Xét \(P=\dfrac{4x}{x-4}\left(4< x< 8\right)\) thì:

Để \(P\in Z\) khi \(\dfrac{4x-16+16}{x-4}\in Z\) <=> \(4+\dfrac{16}{x-4}\in Z\)

=> \(x-4\inƯ\left(16\right)\) mà \(0< x-4< 4\)

=> \(x-4=2\) => \(x=6\)

Vậy \(x\in\left\{6;8;20;68\right\}\) thì \(P\in Z\)

P/s: Vì bài này dài nên mk lm khá tắt, ko hiểu cứ hỏi!

1, ĐKXĐ: x\(\ge0\);x\(\ne1\)

Rút gọn P với \(x\ge0;x\ne1\)ta có

P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)

\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)

=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)

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

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

2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:

P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)

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

=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)

\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3

a: ĐKXĐ: \(x\ge1\)

b: ĐKXĐ: \(x< 0\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)

3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)

4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

5) ĐKXĐ: 

+) \(-x^2+6x+16\ge0\)

\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)

\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)

\(\Leftrightarrow-2\le x\le8\)

+) \(3x^2\ne0\Leftrightarrow x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)

\(đk:\left\{{}\begin{matrix}\Delta\ge0\\0< x1\le x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5^2-4\left(-m^2+m+6\right)\ge0\\\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-4m+1=\left(2m-1\right)^2\ge0\left(đúng\right)\\\left\{{}\begin{matrix}5>0đúng\\-m^2+m+6>0\Leftrightarrow-2< m< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-2< m< 3\)

\(\Rightarrow\dfrac{1}{\sqrt{x1}}+\dfrac{1}{\sqrt{x2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{\sqrt{x1}+\sqrt{x2}}{\sqrt{x1x2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x1+x2+2\sqrt{x1x2}}{x1x2}=\dfrac{9}{4}\Leftrightarrow\dfrac{5+2\sqrt{-m^2+m+6}}{-m^2+m+6}=\dfrac{9}{4}\)

\(đặt::\sqrt{-m^2+m+6}=t\ge0\Rightarrow\dfrac{5+2t}{t^2}=\dfrac{9}{4}\)

\(\Rightarrow9t^2-8t-20=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{10}{9}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-m^2+m+6}=2\Leftrightarrow\left[{}\begin{matrix}m=2\left(tm\right)\\m=-1\left(tm\right)\end{matrix}\right.\)

1.

\(y'=12x+\dfrac{4}{x^2}\)

2.

\(y'=\dfrac{3}{\left(-x+1\right)^2}\)

3.

\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)

4.

\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)

\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)

5.

\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)

6.

\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)