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13 tháng 12 2021
\(a,=\left(2\sqrt{6}-4\sqrt{3}\right)\sqrt{6}+12\sqrt{2}=12-12\sqrt{2}+12\sqrt{2}=12\\ b,=\dfrac{6\left(3-\sqrt{3}\right)}{6}+\sqrt{3}=3-\sqrt{3}+\sqrt{3}=3\)
26 tháng 12 2021
a: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
14 tháng 12 2021
\(M=\dfrac{3}{2}\cdot4\sqrt{2x}-\dfrac{1}{3}\cdot3\sqrt{2x}+\dfrac{2}{5}\cdot5\sqrt{2x}-4\sqrt{2x}=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
14 tháng 12 2021
\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
6 tháng 2 2022
ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\)
\(=\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{\left(\sqrt{x}+3\right)^2}\)
\(=\dfrac{3\sqrt{x}+9-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)
\(=\dfrac{\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
Vậy...
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2021
a.
\(2x-x^2+7=-\left(x^2-2x+1\right)+8=-\left(x-1\right)^2+8\le8\)
\(\Rightarrow2+\sqrt{2x-x^2+7}\le2+\sqrt{8}=2+2\sqrt{2}\)
\(\Rightarrow\dfrac{3}{2+\sqrt{2x-x^2+7}}\ge\dfrac{3}{2+2\sqrt{2}}=\dfrac{3\sqrt{2}-3}{2}\)
\(A_{min}=\dfrac{3\sqrt{2}-3}{2}\) khi \(x=1\)
b. ĐKXĐ: \(x\le1\)
\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}-\dfrac{1}{2}-1\right)\)
\(B=-\left(1-x-\sqrt{2\left(1-x\right)}+\dfrac{1}{2}\right)+\dfrac{3}{2}\)
\(B=-\left(\sqrt{1-x}-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}\le\dfrac{3}{2}\)
\(B_{max}=\dfrac{3}{2}\) khi\(x=\dfrac{1}{2}\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
25 tháng 9 2021
1) ĐKXĐ: \(16x^2-25\ge0\)
\(\Leftrightarrow x^2\ge\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{4}\\x\le-\dfrac{5}{4}\end{matrix}\right.\)
2) ĐKXĐ: \(4x^2-49\ge0\Leftrightarrow x^2\ge\dfrac{49}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{7}{2}\\x\le-\dfrac{7}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(8-x^2\ge0\Leftrightarrow x^2\le8\)
\(\Leftrightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)
4) ĐKXĐ: \(x^2-12\ge0\Leftrightarrow x^2\ge12\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\sqrt{3}\\x\le-2\sqrt{3}\end{matrix}\right.\)
5) ĐKXĐ: \(x^2+4\ge0\left(đúng\forall x\right)\)
\(\Leftrightarrow2\sqrt{x-4}=5\left(x\ge4\right)\\ \Leftrightarrow\sqrt{x-4}=\dfrac{5}{2}\\ \Leftrightarrow x-4=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)
cảm ơn nha