K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1: ĐKXĐ: a>=0; a<>1

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

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

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

2: Để A là số nguyên thì \(\sqrt{a}-1⋮\sqrt{a}+1\)

=>\(\sqrt{a}+1-2⋮\sqrt{a}+1\)

=>\(-2⋮\sqrt{a}+1\)

=>\(\sqrt{a}+1\in\left\{1;-1;2;-2\right\}\)

=>\(\sqrt{a}\in\left\{0;-2;1;-3\right\}\)

=>\(\sqrt{a}\in\left\{0;1\right\}\)

=>\(a\in\left\{0;1\right\}\)

Kết hợp ĐKXĐ, ta được: a=0

18 tháng 12 2023

a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a\ne1\end{matrix}\right.\)

b: Sửa đề: \(C=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)

\(=\left[1:\dfrac{a+\sqrt{1}-\sqrt{a}}{\sqrt{a}+1}\right]\cdot\left[\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)

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

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

c: Để C là số nguyên thì \(a-1⋮a+1\)

=>\(a+1-2⋮a+1\)

=>\(-2⋮a+1\)

=>\(a+1\in\left\{1;-1;2;-2\right\}\)

=>\(a\in\left\{0;-2;1;-3\right\}\)

Kết hợp ĐKXĐ, ta được: a=0

AH
Akai Haruma
Giáo viên
1 tháng 4 2021

Lời giải:

ĐK: $x>0; a\neq 1; a\neq 4$

a) 

$M=\frac{\sqrt{a}-(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}-1)}:\frac{(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}-2)(\sqrt{a}+2)}{(\sqrt{a}-2)(\sqrt{a}-1)}$

$=\frac{1}{\sqrt{a}(\sqrt{a}-1)}:\frac{3}{(\sqrt{a}-2)(\sqrt{a}-1)}=\frac{1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}-2)(\sqrt{a}-1)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}$

b) 

$M>\frac{-1}{2}\Leftrightarrow \frac{\sqrt{a}-2}{3\sqrt{a}}+\frac{1}{2}>0$

$\Leftrightarrow \frac{5\sqrt{a}-4}{6\sqrt{a}}>0$

$\Leftrightarrow 5\sqrt{a}-4>0$

$\Leftrightarrow a>\frac{16}{25}$

Kết hợp với ĐKXĐ thì $a>\frac{16}{25}; a\neq 1; a\neq 4$

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

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

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

Để A nguyên thì \(\sqrt{x}-1\in\left\{-1;1;2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;3\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

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

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

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

b: Để P<1 thì P-1<0

\(\Leftrightarrow\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)

hay 0<a<1

26 tháng 12 2021

a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

27 tháng 12 2021

\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

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

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

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

\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)

c) Để M=-4 thì \(-2\sqrt{a}=-4\)

\(\Leftrightarrow\sqrt{a}=2\)

hay a=4(thỏa ĐK)

22 tháng 12 2020

Bài 1: 

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

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

b) Để Q dương thì \(\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)

mà \(3\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ

nên \(\sqrt{a}-2>0\)

\(\Leftrightarrow\sqrt{a}>2\)

hay a>4

Kết hợp ĐKXĐ,ta được: a>4

Vậy: Để Q dương thì a>4