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a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b,\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)
c, \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x=3x-2x=x\)
d, câu này sai đề rồi , nếu sửa lại phải như này :
\(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}=x-4+4-x=0\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
c) \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x\) = \(\left|3x\right|-2x=-3x-2x\) (x < 0)
= \(-5x\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) ( \(x>4\))
= \(2x-8\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
a, Ta có : \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}\times1+1^2=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}\)
Ta có : \(\sqrt{3}>\sqrt{1}\)(vì 3>1)
\(\Leftrightarrow\sqrt{3}>1\Leftrightarrow\sqrt{3}-1>0\Rightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) (vì \(x>4\))
= \(2x-8\)
a) \(\frac{\sqrt{x}-2}{\sqrt{x}+2}< 0\left(Đk:x\ge0\right)\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
(\(\sqrt{x}+2>0\forall x\ge0\))
Vậy với \(0\le x< 4\)thì ...........
b) \(\frac{3}{\sqrt{x}-5}>0\left(Đk:x\ge0,x\ne25\right)\Leftrightarrow\sqrt{x}-5>0\Leftrightarrow\sqrt{x}>5\Leftrightarrow x>25\)
Vậy với x>25 thì ................
c)
\(\frac{\sqrt{x}-1}{\sqrt{x}-2}< 1\left(ĐK:x\ge0;x\ne4\right)\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}< 0\\ \Leftrightarrow\frac{1}{\sqrt{x}-2}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)
Vậy với \(0\le x< 4\)thì .................
Cái cuối để mk nghĩ đã =v=
1.Ta co:
\(\text{ }\sqrt{5x^2+10x+9}=\sqrt{5\left(x+1\right)^2+4}\ge2\)
\(\sqrt{2x^2+4x+3}=\sqrt{2\left(x+1\right)^2+1}\ge1\)
\(\Rightarrow A=\sqrt{5x^2+10x+9}+\sqrt{2x^2+4x+3}\ge2+1=3\)
Dau '=' xay ra khi \(x=-1\)
Vay \(A_{min}=3\)khi \(x=-1\)
d) \(\sqrt{x+1}+2=0\)( ko tìm đc )
e) \(9x^2=4\Leftrightarrow x^2=\frac{4}{9}\Leftrightarrow x=\pm\sqrt{\frac{4}{9}}\)
g) \(2x^2=\frac{9}{50}\Leftrightarrow x^2=\frac{9}{100}\Leftrightarrow x=\pm\sqrt{\frac{9}{100}}\)
z) \(3-2x=1\Leftrightarrow2x=2\Leftrightarrow x=1\)
y) \(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a)
\(2\sqrt{x}< 16\\ \Leftrightarrow\sqrt{x}< 8\\ \Leftrightarrow x< 64\)
Vậy...
b)
\(3\sqrt{x}+2=0\\ \Leftrightarrow3\sqrt{x}=-2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{3}\)
Nhận xét:
\(\sqrt{x}\) xác định khi và chỉ khi \(\sqrt{x}>0\)
Mà \(-\dfrac{2}{3}< 0\) nên:
Không có giá trị x thoả mãn
Vậy...
c)
\(\sqrt{1-2x^2}=x-1\)
Nhận xét:
\(\sqrt{1-2x^2}\) xác định khi và chỉ khi \(\sqrt{1-2x^2}>0\)
Suy ra:
\(x-1>0\)
\(\Leftrightarrow x>1\)
\(\Leftrightarrow1-2x^2< 0\) (vô lí)
Vậy...
d)
\(2\sqrt{x}-6>0\\ \Leftrightarrow2\sqrt{x}>6\\ \Leftrightarrow\sqrt{x}>3\\ \Leftrightarrow x>9\)
Vậy...