Question

[30.Nɠυұễɳ Tɦàɲɦ Pɦúƈ ヅ]

Bài 1: Tính

a) \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\) 

b) \(\left(\sqrt{14}-\sqrt{10}\right)\sqrt{6+\sqrt{35}}\) 

c) \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\) 

Bài 2: Cho biểu thức 

A = \(\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\) 

a) Rút gọn A

b) Tìm x để A = 2

c) Tìm các số nguyên của x để A ∈ Z

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)

\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)

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

b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)

c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)

\(=\sqrt{3}-\sqrt{3}-1=-1\)

Bài 2:

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

\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

b: A=2

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

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

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

=>x=4(nhận)

c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\inƯ\left(1\right)\)

=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{2;0\right\}\)

=>\(x\in\left\{4;0\right\}\)