\(\dfrac{x-2\sqrt{x+5}}{\sqrt{2x^2+1}}\) có nghĩa khi

\(\left\{{}\begin{matrix}x+5\ge0\\2x^2+1>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\2x^2+1>0\forall x\in R\end{matrix}\right.\\ \Rightarrow x\ge-5\)

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

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

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

a: ĐKXĐ: \(x\ge\dfrac{1}{3}\)

b: ĐKXĐ: \(x< \dfrac{15}{2}\)

c: ĐKXĐ: \(x\le0\)

\(A=\) \(\dfrac{x+2}{x-5}\)

\(=\dfrac{\left(x-5\right)+7}{x-5}\)

\(=1+\dfrac{7}{x-5}\)

để \(\dfrac{7}{x-5}\) ∈Z thì 7⋮x-5

⇒x-5∈\(\left(^+_-1,^+_-7\right)\)

Còn lại thì bạn tự tính nha

1) Ta có: \(P=\dfrac{1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-\sqrt{x}}{x+1}\left(\dfrac{1}{x-2\sqrt{x}+1}+\dfrac{1}{1-x}\right)\)

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

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

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

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

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

Để \(P=-\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}-1}{x+1}=\dfrac{-2}{5}\)

\(\Leftrightarrow-2x-2=5\sqrt{x}-5\)

\(\Leftrightarrow-2x-2-5\sqrt{x}+5=0\)

\(\Leftrightarrow-2x-5\sqrt{x}+3=0\)

\(\Leftrightarrow-2x-6\sqrt{x}+\sqrt{x}+3=0\)

\(\Leftrightarrow-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}+3\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(-2\sqrt{x}+1\right)=0\)

\(\Leftrightarrow-2\sqrt{x}+1=0\)

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

\(\Leftrightarrow x=\dfrac{1}{4}\)(thỏa ĐK)

a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:

\(\left(x-3\right)\left(7-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow3\le x\le7\)

c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:

\(\left(2x-5\right)\left(2x+5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow x>2\)

a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

1: ĐKXĐ: 3x^2-x+2>=0

=>x thuộc R

2: ĐKXĐ: x>=0 và căn x-1<>0 và 2-căn x<>0 và 2x+1>0 và x<>0

=>x>0 và x<>1 và x<>4

Lời giải:

$B=\frac{(x+1)+1}{x+1}=1+\frac{1}{x+1}$

Để $B$ nguyên thì $\frac{1}{x+1}$ nguyên. 

Với $x$ nguyên, để $\frac{1}{x+1}$ nguyên thì $1\vdots x+1$

$\Rightarrow x+1\in\left\{\pm 1\right\}$

$\Rightarrow x\in\left\{0;-2\right\}$

Với $x$ nguyên, để $\frac{5}{2x+7}$ nguyên thì:

$5\vdots 2x+7$

$\Rightarrow 2x+7\in\left\{\pm 1;\pm 5\right\}$

$\Rightarrow x\in\left\{-3;-4;-1;-6\right\}$

B=\(\dfrac{x+2}{x+1}=1\dfrac{1}{x+1}\)(x khác -1)

=> Để B nguyên thì 1 chia hết cho x+1

=> x+1 ∈Ư(1)={1,-1}

Vậy để B nguyên thì x∈{0,-2}

C=\(\dfrac{5}{2x+7}\)(x khác -7/2)

Để C nguyên thì 5 chia hết cho 2x+7

=>2x+7∈Ư(5)={1,-1,5,-5}

Để C nguyên thì x∈{-3,-4,-1,-6}

Để B=\(\dfrac{x+2}{x+1}\) là số nguyên thì x+2 ⋮ x+1

x+2 ⋮ x+1

⇒x+1+1 ⋮ x+1

⇒1 ⋮ x+1

Ta có bảng:

x+1=-1 ➜x=-2 

x+1=1 ➜x=0

Vậy x ∈ {-2;0}

Để C= \(\dfrac{5}{2x+7}\) là số nguyên thì 5 ⋮ 2x+7 

5 ⋮ 2x+7

⇒2x+7 ∈ Ư(5)={-5;-1;1;5}

Ta có bảng giá trị:

2x+7=-5 ➜x=-6

2x+7=-1 ➜x=-4

2x+7=1 ➜x=-3

2x+7=5 ➜x=-1

Vậy x ∈ {-6;-4;-3;-1}

Chúc bạn học tốt!

a: Để D là số nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)