(√24 - √48) .√6 +12√2
\(\dfrac{6}{\sqrt{3}+3}+\dfrac{3}{\sqrt{3}}\)
giải giúp mk chi tiết hai câu này vớiiiiii ạ
Rút gọn biểu thức
M=\(\dfrac{3}{2}\sqrt{32x}-\dfrac{1}{3}\sqrt{18x}+\dfrac{2}{5}\sqrt{50x}-4\sqrt{2x}\) (x ≥ 0)
giải chi tiết giúp mk vớiiiiii ạ
\(M=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
\(\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\) với x>0; x khác 4
ai giải chi tiết giúp mk bài này với khó quá
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...
f)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
g)\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0,4}\right)\)
giải chi tiết cụ thể giúp mk với ạ
1.)\(\sqrt{\left(\sqrt{3}-3\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
2)\(\dfrac{1}{\sqrt{5}-2}\) +\(\dfrac{\sqrt{10}-\sqrt{5}}{1-\sqrt{2}}\)
3)\(\dfrac{\sqrt{2}}{\sqrt{3}-1}\) -\(\sqrt{\dfrac{3}{2}}\)
giải chi tiết giúp mk vớiiiii ạ
1)\(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)
2: \(=\sqrt{5}+2-\sqrt{5}=2\)
M= \(\dfrac{3}{2}\sqrt{32x}-\dfrac{1}{3}\sqrt{18x}+\dfrac{2}{5}\sqrt{50x}-4\sqrt{2x}\) (x lớn hơn hoặc bằng 0)
giải chi tiết giúp mk vớiiiii ạ
\(M=\dfrac{3}{2}\cdot4\sqrt{2x}-\dfrac{1}{3}\cdot3\sqrt{2x}+\dfrac{2}{5}\cdot5\sqrt{2x}-4\sqrt{2x}=6\sqrt{2x}-\sqrt{2x}+2\sqrt{2x}-4\sqrt{2x}=3\sqrt{2x}\)
bài 1:giải phương trình
a)\(\sqrt{9x^2+12x+4}-4\) = 0
b)\(3\sqrt{x+3}-\sqrt{x-5}\) = 0
c)\(x-7+\sqrt{x-1}\) = 0
giải cụ thể chi tiết giúp mk vớiiiiii ạ
a: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
Bài 1: Thực hiện phép tính:
a, \(\left(\sqrt{24}-\sqrt{48}-\sqrt{6}\right)\sqrt{6}+12\sqrt{2}\)
b, \(\left(\sqrt{\dfrac{1}{5}}-\sqrt{\dfrac{16}{5}}+\sqrt{5}\right):\sqrt{20}\)
c, \(\sqrt{21+3\sqrt{48}}-\sqrt{21-3\sqrt{48}}\)
Bài 2: Giải các phương trình sau:
a, \(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\)
b, \(\sqrt{9x^2+12x +4}=4x\)
c, \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\)
GIÚP MIK VỚIIII
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
mn cs thể giải chi tiết đc ko ạ
\(A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 4$
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-1\right]:\left[\frac{(3-\sqrt{x})(3+\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+3)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right]\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{-(\sqrt{x}-2)}=\frac{3}{\sqrt{x}-2}\)
Tìm điều kiện có nghĩa:
1) \(\sqrt{x^2+2x-3}\)
2) \(\sqrt{2x^2+5x+3}\)
3) \(\sqrt{\dfrac{4}{x-1}}\)
4) \(\sqrt{\dfrac{-1}{x-3}}\)
5) \(\sqrt{\dfrac{-3}{x+2}}\)
6) \(\sqrt{\dfrac{1}{2a-1}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)
2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)
4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)
5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)
6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)