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a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)
c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)
d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
Ta có:
x = \(\frac{1}{2}\)\(\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
= \(\frac{1}{2}\)\(\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{1}}\)
= \(\frac{1}{2}\)(\(\sqrt{2}\)-1)
=> 2x = \(\sqrt{2}\)-1
=> (2x)2= ( \(\sqrt{2}\)-1)2
=> 4x2= 2-2\(\sqrt{2}\)+1
=> 4x2= -2( \(\sqrt{2}\)-1)+1
=> 4x2= -4x +1 => 4x2+4x-1=0
Lại có:
A1= (\(4x^5\)+\(4x^4\)- \(x^3\)+1)19
= [ x3( 4x2+4x-1) +1]19
=1
A2=( \(\sqrt{4x^5+4x^4-5x^3+5x+3}\))3
= (\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\))3
= 23=8
A3= \(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\)
= \(\sqrt{2}\)- \(\sqrt{2}\)\(\sqrt{1-\sqrt{2}}\)
Cộng 3 số vào ta được A
a)\(\sqrt{\frac{2x-3}{x-1}}=2\RightarrowĐk:\frac{2x-3}{x-1}\ge0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{3}{2}\\x< 1\end{array}\right.\)
\(\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4\left(x-1\right)\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)(nhận)
b)\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\RightarrowĐk:\begin{cases}2x-3\ge0\\x-1>0\end{cases}\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow\sqrt{2x-3}=2\sqrt{x-1}\)
\(\Leftrightarrow2x-3=4x-4\)\(\Leftrightarrow x=\frac{1}{2}\)(loại)
c)\(\sqrt{\frac{4x+3}{x+1}}=3\RightarrowĐk:\frac{4x+3}{x+1}\ge0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\ge\frac{-3}{4}\\x< -1\end{array}\right.\)
\(\sqrt{\frac{4x+3}{x+1}}=3\Rightarrow\frac{4x+3}{x+1}=9\)
\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow5x=-6\Leftrightarrow x=\frac{-6}{5}\)(nhận)
c)\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\RightarrowĐk:\begin{cases}4x+3\ge0\\x+1>0\end{cases}\)
\(\Rightarrow x\ge\frac{-3}{4}\)
\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Rightarrow\sqrt{4x+3}=3\sqrt{x+1}\)
\(\Leftrightarrow4x+3=9\left(x+1\right)\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow x=\frac{-6}{5}\)(loại)