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a) ĐKXĐ: \(5x-7\ge0\) \(\Leftrightarrow\)\(x\ge\frac{7}{5}\)
b) ĐKXĐ: \(2x^2+x\ge0\)\(\Leftrightarrow\) \(x\left(2x+1\right)\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x\ge0\\x\le-\frac{1}{2}\end{cases}}\)
c) ĐKXĐ: \(4-7x\ge0\)\(\Leftrightarrow\)\(x\le\frac{4}{7}\)
d) ĐKXĐ: \(x^3+x\ge0\) \(\Leftrightarrow\)\(x\left(x^2+1\right)\ge0\)\(\Leftrightarrow\)\(x\ge0\)
e) ĐKXĐ: \(\frac{x-5}{2x+1}\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x\ge5\\x< -\frac{1}{2}\end{cases}}\)
f) ĐKXĐ: \(\frac{3-2x}{3x-2}\ge0\) \(\Leftrightarrow\)\(\frac{2}{3}< x\le\frac{3}{2}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)
\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)
\(\Rightarrow3x+2=2\left(x+2\right)\)
\(\Rightarrow3x+2=2x+4\)
\(\Rightarrow3x-2x=4-2\)
\(\Rightarrow x=2\)
\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Rightarrow2\sqrt{x-2}=4\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)
\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)
\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)
\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)
\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)
\(\Rightarrow2x^2+7x=0\)
\(\Rightarrow x\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)
\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)
\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)
\(\Rightarrow x=1\)
a/ 2x-x2>0
\(\Leftrightarrow\) x(2-x)>0
\(\Leftrightarrow\) 0<x<2
b/ \(\left\{{}\begin{matrix}x-3>0\\5-x>0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x>3\\x< 5\end{matrix}\right.\)\(\Leftrightarrow\) 3<x<5
c/ x2-5x+6>0
\(\Leftrightarrow\) (x-3)(x-2)>0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x>3\\x< 2\end{matrix}\right.\)
d/ \(\left\{{}\begin{matrix}6x-1>0\\x+3>0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x>\frac{1}{6}\\x>-3\end{matrix}\right.\)
\(\Leftrightarrow\) x > \(\frac{1}{6}\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Lời giải:
a)
\(\left\{\begin{matrix} x\geq 0\\ 3-\sqrt{x}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\leq 9\end{matrix}\right.\Leftrightarrow 0\leq x\leq 9\)
b)
\(\left\{\begin{matrix} x-1\geq 0\\ 2-\sqrt{x-1}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x-1\leq 4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 5\end{matrix}\right.\)
\(\Leftrightarrow 1\leq x\leq 5\)
c)
\(-7+3x>0\Leftrightarrow x>\frac{7}{3}\)
d)
\(\left\{\begin{matrix} x-1\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x< 5\end{matrix}\right.\Leftrightarrow 1\leq x< 5\)
e) \(x\in\mathbb{R}\)
f) \(\left\{\begin{matrix} 2-x>0\\ x-5\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x< 2\\ x\geq 5\end{matrix}\right.\) (vô lý)
Do đó không tồn tại $x$ để hàm số tồn tại
g)
\(\left[\begin{matrix} \left\{\begin{matrix} 3x-6-2x\geq 0\\ 1-x>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x-6-2x\leq 0\\ 1-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq 6\\ x< 1\end{matrix}\right.(\text{vô lý})\\ \left\{\begin{matrix} x\leq 6\\ x>1 \end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow 1< x\leq 6\)