a/ ĐKXĐ : \(-2x+3\ge0\)

\(\Leftrightarrow x\le\dfrac{3}{2}\)

b/ ĐKXĐ : \(3x+4\ge0\)

\(\Leftrightarrow x\ge-\dfrac{4}{3}\)

c/ Căn thức \(\sqrt{1+x^2}\) luôn được xác định với mọi x

d/ ĐKXĐ : \(-\dfrac{3}{3x+5}\ge0\)

\(\Leftrightarrow3x+5< 0\)

\(\Leftrightarrow x< -\dfrac{5}{3}\)

e/ ĐKXĐ : \(\dfrac{2}{x}\ge0\Leftrightarrow x>0\)

P.s : không chắc lắm á!

a) \(\sqrt{3x-4}\) xác định \(\Leftrightarrow3x-4\ge0\Leftrightarrow3x\ge4\Leftrightarrow x\ge\dfrac{4}{3}\)

b) \(\dfrac{1}{\sqrt{x-4}}\) xác định \(\Leftrightarrow x-4>0\Leftrightarrow x>4\)

a, đkxđ : x >= 4/3 

b, đkxđ : x > 4 

a: ĐKXĐ: \(\dfrac{x-1}{5-x}\ge0\)

\(\Leftrightarrow\dfrac{x-1}{x-5}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-5< 0\end{matrix}\right.\Leftrightarrow1\le x< 5\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>3\\x< 2\end{matrix}\right.\)

Sửa đề: \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{x-1}\)

a: ĐKXĐ: x>=0; x<>1

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

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

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

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

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

a) ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

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

\(A=\dfrac{-x-4\sqrt{x}+1}{x-1}\)

a) ĐK: x ≥ 2

\(\sqrt{3x-6}=3\)

\(\Leftrightarrow3x-6=9\)

<=> 3x = 15

<=> x = 5

Vậy:....

b) ĐK: 5x - 16 ≥ 0

<=> 5x ≥ 16

<=> x ≥ 16/5

\(\sqrt{5x-16}=2\)

<=> 5x - 16 = 4

<=> 5x = 20

<=> x = 4

c) ĐK: \(x^2-4x+3\ne0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)

bình phương hai vế ta được:

a)điều kiện của x:x≥2

3x-6=9 <=> x=5(nhận)

b)ĐK: x≥16/5

5x-16=4 <=>x=4(nhận)

c) ta có: \(\dfrac{2x-3}{\left(x-2\right)^2-1}\)= \(\dfrac{2x-3}{\left(x-3\right)\left(x-1\right)}\)

ĐKXĐ: x≠3 ;x≠1

a,\(\sqrt{3x-6}=3\) (với x\(\ge\)2)

=>\(\left(\sqrt{3x-6}\right)^2=3^2\)

<=>\(3x-6=9\)<=>\(3x=9+6\)<=>x=\(\dfrac{15}{3}\)=5(thỏa mãn)

b,\(\sqrt{5x-16}=2\) (với x\(\ge\)16/5)

=>\(\left(\sqrt{5x-16}\right)^2=2^2\)<=>\(5x-16=4< =>5x=20< =>x=4\)(thỏa mãn)

c,B xác định khi \(x^2-4x+3\ne0< =>x^2-2.2.x+2^2-1\ne0\)

\(< =>\left(x-2\right)^2-1\ne0\)

\(< =>\left(x-2+1\right)\left(x-2-1\right)\ne\)0

\(< =>\left(x-1\right)\left(x-3\right)\ne0\)

\(< =>\left[{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)

\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

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

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

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

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

a) ĐKXĐ: \(x\ge0;x\ne1\)

b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

Giúp mình với

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

b: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\)

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

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

\(a,ĐK:2-x^2\ge0\Leftrightarrow x^2\le2\Leftrightarrow-\sqrt{2}\le x\le\sqrt{2}\\ b,ĐK:5x^2-3>0\Leftrightarrow x^2>\dfrac{3}{5}\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{\sqrt{15}}{5}\\x< -\dfrac{\sqrt{15}}{5}\end{matrix}\right.\\ c,ĐK:-\left(2x-1\right)^2\ge0\Leftrightarrow x=\dfrac{1}{2}\\ d,ĐK:x^2+x-2>0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)