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a) đk: \(x\ge0;x\ne1\)
b) \(A=\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right)\div\frac{\sqrt{x}-1}{2}\)
\(A=\frac{x+2+\left(\sqrt{x}-1\right)\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\div\frac{\sqrt{x}-1}{2}\)
\(A=\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)
\(A=\frac{2\left(x-2\sqrt{x}+1\right)}{\left(x-2\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{2}{x+\sqrt{x}+1}\)
c) Ta có: \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
=> \(\frac{2}{x+\sqrt{x}+1}>0\left(\forall x\ne1\right)\)
d) Ta chỉ có thể tìm GTLN thôi
Để A đạt GTLN => \(x+\sqrt{x}+1\) phải đạt GTNN
Dấu "=" xảy ra khi: \(x=0\)
Vậy Max(A) = 2 khi x = 0
#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
a) ĐKXĐ: x\(\ne\) 0;4
Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
= \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)= \(\frac{4\sqrt{x}\cdot\left(2+\sqrt{x}\right)}{2+\sqrt{x}}\cdot\frac{-\sqrt{x}}{3-\sqrt{x}}\)=\(\frac{-4}{3-\sqrt{x}}\)=\(\frac{4}{\sqrt{x}-3}\)
b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)
<=> \(4=3-\sqrt{x}\)
<=> \(\sqrt{x}=-1\) (vô lí)
Vậy ko tìm được x.
c) Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)9
A = \(-\frac{1}{\sqrt{x}-3}\) => -2A = \(\frac{2}{\sqrt{x}-3}\)
Để -2A thuộc Z <=> \(2⋮\sqrt{x}-3\)
<=> \(\sqrt{x}-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng:
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 |
| x | 8 | 4 (ktm) | 25 | 1 |
Vậy ....
a/A\(=\frac{x+2}{x-\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+1}-\frac{1-\sqrt{x}}{\sqrt{x}-2}\)
\(=\frac{x+2-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2-2x+4\sqrt{x}-1+x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
Thay x=16 vào A ta có: A\(=\frac{3}{2}\)
b/ B= \(1-\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
\(\frac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}=\frac{1}{\sqrt{x}-2}\)
=>C=\(\frac{4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{1}{\sqrt{x}-2}\)=\(\frac{4\sqrt{x}-1}{\sqrt{x}+1}\)
c/Để C thuộc Z thì \(\frac{4\sqrt{x}-1}{\sqrt{x}+1}\) thuộc Z
C\(=\text{}\frac{4\sqrt{x}-1}{\sqrt{x}+1}=\frac{4\sqrt{x}+4}{\sqrt{x}+1}-\frac{5}{\sqrt{x}+1}=4-\frac{5}{\sqrt{x}+1}\)
=> \(5⋮\left(\sqrt{x}+1\right) \Leftrightarrow\sqrt{x}+1\in\left\{-5;-1;1;5\right\}\)
Nhận xét: \(\sqrt{x}+1\ge1\)
\(\Rightarrow\sqrt{x}+1\in\left\{1;5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;4\right\} \Leftrightarrow x\in\left\{0;16\right\}\)
Vậy \(x\in\left\{0;16\right\}\) thì C thuộc Z
Chúc bạn học tốt!
cảm ơn bạn nhiều <3