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a: \(=4x-4x\sqrt{2}-2x\sqrt{2}+2x=6x-6x\sqrt{2}\)
b: \(=6x-4\sqrt{xy}+3\sqrt{xy}-2y=6x-\sqrt{xy}-2y\)
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) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
a) Đk: x \(\ge\) 5
\(\sqrt{x-5}-\frac{x-14}{3x+\sqrt{x-5}}=3\)
\(\sqrt{x-5}\left(3+\sqrt{x-5}\right)-\frac{x-14}{3\sqrt{x-3}}\left(3+\sqrt{x-5}\right)=3\left(3+\sqrt{x-5}\right)\)
\(\sqrt{x-5}\left(3+\sqrt{x-5}\right)-\left(x-14\right)=3\left(3+\sqrt{x-5}\right)\)
\(3\sqrt{x-5}+9-\left(3\sqrt{x-5}+9\right)=9+3\sqrt{x-5}-\left(3\sqrt{x-5}+9\right)\)
=> Luôn đúng với x \(\ge\) 5
chúc bạn học tốt
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 > \(\frac{1}{6}\)
Lời giải:
Bạn cứ nhớ công thức $\sqrt{x^2}=|x|$, rồi dùng điều kiện đề bài để phá dấu trị tuyệt đối là được
a)
\(\sqrt{16a^2}-5a=\sqrt{(4a)^2}-5a=|4a|-5a=4a-5a=-a\)
b)
\(3x+2-\sqrt{9x^2+6x+1}=3x+2-\sqrt{(3x)^2+2.3x.1+1^2}\)
\(=3x+2-\sqrt{(3x+1)^2}=3x+2-|3x+1|=3x+2-(3x+1)=1\)
c)
\(\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+1+2.\sqrt{7}.\sqrt{1}}-\sqrt{7}\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{7}=|\sqrt{7}+1|-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
d)
\(\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{13+1-2\sqrt{13}}+\sqrt{13+1+2\sqrt{13}}\)
\(=\sqrt{(\sqrt{13}-1)^2}+\sqrt{(\sqrt{13}+1)^2}=|\sqrt{13}-1|+|\sqrt{13}+1|\)
\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)
e)
\(2x-\sqrt{4x^2-4x+1}=2x-\sqrt{(2x-1)^2}=2x-|2x-1|=2x-(2x-1)=1\)
g)
\(|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=|x-2|+\frac{\sqrt{(x-2)^2}}{x-2}=|x-2|+\frac{|x-2|}{x-2}\)
\(=(x-2)+\frac{(x-2)}{x-2}=x-2+1=x-1\)
a: \(=1-\left(\sqrt{x}\right)^3=1-x\sqrt{x}\)
b: \(=\left(\sqrt{x}\right)^3+2^3=x\sqrt{x}+8\)
c: \(=\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3=x\sqrt{x}-y\sqrt{y}\)
d: \(=x^3+\left(\sqrt{y}\right)^3=x^3+y\sqrt{y}\)
Biểu thức Trong căn xác định => Dương hết :)))