G
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
6 tháng 10 2020
a) đk: \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(P=\left(\frac{3x-\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right)\div\frac{1}{x-1}\)
\(P=\frac{3x-3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(x-1\right)\)
\(P=\frac{3x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)
\(P=\frac{\left(3\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
15 tháng 8 2020
ĐKXĐ: x \(\ge\)0; x \(\ne\)1
a) P = \(\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
P = \(\left(\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{5}{x+2\sqrt{x}-\sqrt{x}-2}\right):\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)
P = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) P = \(\frac{1}{\sqrt{x}}\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)
=> \(\sqrt{x}\left(2\sqrt{x}+1\right)-\sqrt{x}-1=0\)
<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)
<=> \(x=\frac{1}{2}\)(tm)
c)Với đk: x \(\ge\)0 và x \(\ne\)1
\(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))
<=> \(x-1-2\sqrt{x-1}+1=0\)
<=> \(\left(\sqrt{x-1}-1\right)^2=0\)
<=> \(\sqrt{x-1}-1=0\)
<=> \(\sqrt{x-1}=1\)
<=> \(\left(\sqrt{x-1}\right)^2=1\)
<=> \(\left|x-1\right|=1\)
<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Với x = 2 => P = \(\frac{2\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{4-2\sqrt{2}+\sqrt{2}-1}{2-1}=3-\sqrt{2}\)
15 tháng 8 2020
a) P = \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)(sửa lại)
b) \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\) => \(2x-\sqrt{x}-\sqrt{x}-1=0\)
<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)
<=> \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)
KN
18 tháng 8 2021
a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)
\(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)
\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa
10 tháng 12 2017
Sau khi rút gọn ,ta được A=\(\left(\sqrt{x}+1\right)^2\)\(\Rightarrow\frac{1}{A}=\frac{1}{\left(\sqrt{x}+1\right)^2}\). Để \(\frac{1}{A}\)là số tự nhiên \(\hept{\begin{cases}\left(\sqrt{x}+1\right)^2>0\\\left(\sqrt{x}+1\right)^2\in U\left(1\right)\end{cases}}\) \(\Rightarrow x=0\)( thỏa mãn ĐK).
11 tháng 7 2019
- a,de bieu thuc 2x^2+3x-1 /2x+1 nguyen thi2x+1 phai khac 0suy ra 2x khac -1suy ra x khac -1/2b,de bieu thuc 2x^2+4x-1/2x+1 nguyen thi 2x+1 khac 0suy ra 2x khac -1suy ra x khac -1/2c,de bieu thuc 2x^2+3/x-2 nguyen thi x-2 khac 0suy ra x khac 2
28 tháng 6 2019
\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)
\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)
28 tháng 6 2019
b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)
\(2\sqrt{x}-1>0\);\(4x>0\)
\(\Rightarrow x>0\)thì \(A>A^2\)