b: ĐKXĐ: \(x\in R\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\)

\(D=\dfrac{7-x}{x-4}=\dfrac{-\left(x-7\right)}{x-4}=\dfrac{-\left(x-4\right)+3}{x-4}=-1+\dfrac{3}{x-4}\)

Để D đạt giá trị nguyên thì:

3 ⋮ x - 4

=> x - 4 ∈ Ư (3)

=> x - 4 ∈ {1; -1; 3; -3}

=> x ∈ {5; 3; 7; 1}

Vậy với x ∈ {5; 3; 7; 1} thì D nhận giá trị nguyên

Tìm tất cả các giá trị nguyên của x để biểu thức sau nhận giá trị nguyên:(2x-5)/(x-4)

ĐKXĐ:

\(2x-18\ge0\)

\(\Rightarrow x\ge9\)

ĐKXĐ: \(x\ge9\)

1: Thay \(x=\dfrac{1}{25}\) vào C, ta được:

\(C=\left(\dfrac{1}{5}+2\right):\left(\dfrac{1}{5}-3\right)=\dfrac{11}{5}:\dfrac{-14}{5}=-\dfrac{11}{14}\)

2: Để C=-2 thì \(\sqrt{x}+2=-2\left(\sqrt{x}-3\right)\)

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

\(\Leftrightarrow3\sqrt{x}=4\)

hay \(x=\dfrac{16}{9}\)

Để \(C=\dfrac{7}{5}\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{7}{5}\)

\(\Leftrightarrow7\sqrt{x}-21=2\sqrt{x}+10\)

\(\Leftrightarrow5\sqrt{x}=31\)

hay \(x=\dfrac{961}{25}\)

A đâu em?

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

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

Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)

b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)

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

\(=2-\sqrt{3}+\dfrac{5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{6-2\sqrt{3}-3\sqrt{3}+3+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)

d: A=2

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

=>\(x+3\sqrt{x}+1=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)

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

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