1.
a, \(\sqrt{18}-2\sqrt{50}+3\sqrt{8}=3\sqrt{2}-2.5\sqrt{2}+3.2\sqrt{2}\)
\(=3\sqrt{2}-10\sqrt{2}+6\sqrt{2}=-\sqrt{2}\)
b, \(\left(\sqrt{7}-\sqrt{3}\right)^2+\sqrt{84}=7+3-2\sqrt{7.3}+\sqrt{84}=10-2\sqrt{21}+2\sqrt{21}=10\)
2.
a, \(\sqrt{\left(2x+3\right)^2}=4\Leftrightarrow\left|2x+3\right|=4\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy x\(\in\left\{\dfrac{1}{2};\dfrac{-7}{2}\right\}\)
b, \(\sqrt{9x}-5\sqrt{x}=6-4\sqrt{x}\left(ĐKXĐ:x\ge0\right)\)
\(\Leftrightarrow3\sqrt{x}-5\sqrt{x}+4\sqrt{x}=6\)
\(\Leftrightarrow2\sqrt{x}=6\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tmĐKXĐ\right)\)
Vậy x = 9
3.
a, ĐKXĐ:\(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
\(Q=\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}:\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(Q=\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\times\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(Q=\dfrac{1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(Q=\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(Q=\dfrac{a-\sqrt{a}-a-2\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{-3\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
. Giúp mình với nhé cảm ơn nhiều!
Mn giúp mình bài 3 với ạ. Mình đang cần gấp lắm. Mình cảm ơn nhiều !
