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: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)

\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)

\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)

\(=10\cdot\dfrac{2}{5}=4\)

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:

\(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt[]{x}+1}\\ DKXD:x\ne1;x\ge0\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ A=\sqrt{x}-1+\sqrt{x}\\ A=2\sqrt{x}+1\)

\(C.A< -1\Leftrightarrow2\sqrt{x}-1< -1\\ \Leftrightarrow2\sqrt{x}< 0\\ \Leftrightarrow x< 0\left(ktmdk\right)\\ =>BPTVN:S=\varnothing\)

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) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

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

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

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)

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:

\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

a) Ta có: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x^2+\sqrt{x}}{x\sqrt{x}+x}\)

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

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

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

\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)

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: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

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: \(0\le x< 1\)

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