HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
a) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{4.3}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}\)\(=\dfrac{\sqrt{6}}{2}\)
c)\(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)
\(\)c)\(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)\)
\(\Leftrightarrow6+6\sqrt{7}-2\sqrt{6}-14\)
\(\Leftrightarrow-8+5\sqrt{6}\)
d)\(\left(\sqrt{3}+2\right)\left(\sqrt{3}-5\right)\)
\(\Leftrightarrow3-5\sqrt{3}+2\sqrt{3}-3\)
\(\Leftrightarrow-3\sqrt{3}\)
1. c)\(\left(\sqrt{6}-2\right)\left(\sqrt{6}+7\right)\)
\(\Leftrightarrow6+7\sqrt{6}-2\sqrt{6}-14\)
\(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=9\\2x-3=9\\x-1=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=6\\x=10\end{matrix}\right.\)
Vậy \(x=\left\{3,5;6;10\right\}\)
\(\sqrt{16x^2}-2x=4x-2x=2x\)
Bạn ơi đề bài là j
a)\(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(A=\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(A=\dfrac{x\sqrt{x}-3-\left(2\sqrt{x}+6\right)-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x\sqrt{x}-3-2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{-2\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}\)
\(\Leftrightarrow\sqrt{x^2-3^2}+\sqrt{x^2-2.3.x+9}=0\)
\(\Leftrightarrow x-3+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow x-3+x-3=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b.\(\sqrt{2}.\sqrt{162}\)
\(=\sqrt{81}\cdot\sqrt{2^2}\)\(=9\cdot2=18\)