\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).

a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)

⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)

Phương trình vô nghiệm

b, x = \(\dfrac{8}{125}\)

a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))

Vì hai vế ko âm, bp 2 vế ta được:

2x² - 3 = 4x - 3

\(\Leftrightarrow\) 2x² = 4x

\(\Leftrightarrow\) x² = 2x

\(\Leftrightarrow\) x² - 2x = 0

\(\Leftrightarrow\) x(x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

Vậy S = {2}

b, \(\sqrt{2x-1}=\sqrt{x-1}\) (x \(\ge\) 1)

Vì hai vế ko âm, bp 2 vế ta được:

2x - 1 = x - 1

\(\Leftrightarrow\) x = 0 (KTM)

Vậy x = \(\varnothing\)

c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)

Vì hai vế ko âm, bp 2 vế ta được:

x² - x - 6 = x - 3

\(\Leftrightarrow\) x² - 2x - 3 = 0

\(\Leftrightarrow\) x² - 3x + x - 3 = 0

\(\Leftrightarrow\) x(x - 3) + (x - 3) = 0

\(\Leftrightarrow\) (x - 3)(x + 1) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy S = {3}

d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))

Vì hai vế ko âm, bp 2 vế ta được:

x² - x = 3x - 5

\(\Leftrightarrow\) x² - 4x + 5 = 0

\(\Leftrightarrow\) x² - 4x + 4 + 1 = 0

\(\Leftrightarrow\) (x - 2)² + 1 = 0

Vì (x - 2)² \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)² + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!

Nguyễn Lê Phước Thịnh nhờ anh xíu ạ

\(c,=2+2\sqrt{3}-\left(2+\sqrt{2}\right)=2\sqrt{3}-\sqrt{2}\\ d,=\sqrt{\left(2x-3\right)^2}-2x+1=\left|2x-3\right|-2x+1\\ =2x-3-2x+1=-2\left(x\ge\dfrac{3}{2}\Leftrightarrow2x-3\ge0\right)\)