Question

Giúp mình với!

Answer 1

Bài 9:

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a) Ta có: \(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

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

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

b) Ta có: \(\left|2x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)

Thay x=4 vào Q, ta được:

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

c) Để Q=3 thì \(\sqrt{x}+1=3\left(\sqrt{x}-1\right)\)

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

\(\Leftrightarrow-2\sqrt{x}=-4\)

hay x=4(thỏa ĐK)

Answer 2

Bài 8:

a) Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

b) Thay \(x=11-6\sqrt{2}\) vào M, ta được:

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

c) Để M=2 thì \(\sqrt{x}+1=2\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow-\sqrt{x}=-7\)

hay x=49

Answer 3

bài 8 còn 3 phần cuối cậu giúp mình với

 

Answer 4

Bài 8:

d) Để M<1 thì M-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

e) Để M nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7\right\}\)

hay \(x\in\left\{1;4;16;25;49\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;16;25;49\right\}\)