Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)
A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)
A = 4
Vậy A không phụ thuộc vào x
Chúc bn học tốt!
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)
\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)
\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)
\(=10\cdot\dfrac{2}{5}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\ge0,x\ne1\)
b) \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
c) \(A=2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\)(vô lý do \(2\sqrt{x}\ge0\forall x\))
Vậy \(S=\varnothing\)
Bài 3:
\(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt[]{x}+1}\\ DKXD:x\ne1;x\ge0\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ A=\sqrt{x}-1+\sqrt{x}\\ A=2\sqrt{x}+1\)
\(C.A< -1\Leftrightarrow2\sqrt{x}-1< -1\\ \Leftrightarrow2\sqrt{x}< 0\\ \Leftrightarrow x< 0\left(ktmdk\right)\\ =>BPTVN:S=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{x^2-x+1}\)
\(=\sqrt{x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}}\)
\(=\sqrt{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Mà: \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
Nên bt luôn có nghĩa
b) \(\dfrac{5}{\sqrt{1-\sqrt{x-1}}}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-1\ge0\\1-\sqrt{x-1}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x-1< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\\x< 2\end{matrix}\right.\Leftrightarrow1\le x< 2\)
c) \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) có nghĩa khi:
\(x\ge0\)
d) \(\dfrac{\sqrt{-3x}}{x^2-1}\) có nghĩa khi:
\(\Leftrightarrow\left\{{}\begin{matrix}-3x\ge0\\x^2-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x\ne\pm1\end{matrix}\right.\)
e) \(\dfrac{2}{\sqrt{x}-2}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-2\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) để căn thức có nghĩa thì \(3x^2+1\ge0\) (luôn đúng) nên căn luôn có nghĩa
b) để căn thức có nghĩa thì \(4x^2-4x+1\ge0\Rightarrow\left(2x-1\right)^2\ge0\) (luôn đúng)
nên căn luôn có nghĩa
c) để căn thức có nghĩa thì \(\dfrac{3}{x+4}\ge0\) mà \(3>0\Rightarrow x+4>0\Rightarrow x>-4\)
h) để căn thức có nghĩa thì \(x^2-4\ge0\Rightarrow x^2\ge4\Rightarrow\left|x\right|\ge2\)
i) để căn thức có nghĩa thì \(\dfrac{2+x}{5-x}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2+x\ge0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2+x\le0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2\le x< 5\\\left\{{}\begin{matrix}x\le-2\\x>5\end{matrix}\right.\left(l\right)\end{matrix}\right.\Rightarrow-2\le x< 5\)
a) ĐKXĐ: \(x\in R\)
b) ĐKXĐ: \(x\in R\)
c) ĐKXĐ: x>-4
h) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}\)
\(=2\sqrt{x}-1\)
b) Để P<1 thì \(2\sqrt{x}< 2\)
\(\Leftrightarrow x< 1\)
Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)
ủa bạn ơi, sao cái dấu bằng đầu tiên đã ra như vậy rồi, mình ko hiểu
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)
\(P=\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}-3}{4-x}\)
\(=\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{4-x}{\sqrt{x}-3}\)
\(=\dfrac{-4\left(4-x\right)}{\left(x-4\right)\left(\sqrt{x}-3\right)}=\dfrac{4}{\sqrt{x}-3}\)
b: P>-1
=>P+1>0
=>\(\dfrac{4}{\sqrt{x}-3}+1>0\)
=>\(\dfrac{4+\sqrt{x}-3}{\sqrt{x}-3}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}>0\)
=>\(\sqrt{x}-3>0\)
=>x>9
a, ĐKXĐ: \(x^2-3\ge0\Rightarrow x^2\ge3\Rightarrow x\ge\sqrt{3}\)
b, \(\left\{{}\begin{matrix}x-2\ne0\\x-2\ge0\end{matrix}\right.\Rightarrow x-2>0\Rightarrow x>2\)
c, \(\left\{{}\begin{matrix}3-2x\ne0\\\dfrac{1}{3-2x}\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x\ne3\\3-2x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{3}{2}\\x< \dfrac{3}{2}\end{matrix}\right.\)
\(\sqrt{x^2-3}\)
ĐKXĐ: x > 1
\(\dfrac{x}{x-2}+\sqrt{x-2}\)
ĐKXĐ: x > 2
\(\sqrt{\dfrac{1}{3-2x^2}}\)
ĐKXĐ: x < 1,224744871 \(\approx\) 1,22