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a: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)
\(=\sqrt{4-2\cdot2\cdot\sqrt2+2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)
\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{\left(3\sqrt2-2\right)^2}\)
\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)
b: \(\sqrt{\left(\sqrt3-\sqrt2\right)^2}+\sqrt2=\sqrt3-\sqrt2+\sqrt2=\sqrt3\)
c: \(3\sqrt5-\sqrt{\left(1-\sqrt5\right)^2}\)
\(=3\sqrt5-\left|1-\sqrt5\right|\)
\(=3\sqrt5-\left(\sqrt5-1\right)=2\sqrt5+1\)
d:Sửa đề: \(\sqrt{17-12\sqrt2}+\sqrt{6+4\sqrt2}\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt2+8}+\sqrt{4+2\cdot2\cdot\sqrt2+2}\)
\(=\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{\left(2+\sqrt2\right)^2}=3-2\sqrt2+2+\sqrt2=5-\sqrt2\)
Câu A=4
Cách giải:
\(\left(5\sqrt{3}+2\sqrt{12}-\sqrt{75}\right):\sqrt{3}\)
\(=\left(5\sqrt{3}+2\sqrt{4\cdot3}-\sqrt{25\cdot3}\right)\)\(:\sqrt{3}\)
\(=\left(5\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\)\(:\sqrt{3}\)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
1/ ĐKXĐ : \(0\le a\ne1\)
2/ \(A=\left(\frac{\sqrt{a}-2}{a-1}-\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}\right).\frac{\left(1-a\right)^2}{2}\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}+2\right)\left(a-1\right)}{\left(a-1\right)\left(\sqrt{a}+1\right)^2}.\frac{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^3}.\frac{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}{2}\)
\(=-\sqrt{a}\left(\sqrt{a}-1\right)\)
3/ \(A=-\sqrt{a}\left(\sqrt{a}-1\right)=-a+\sqrt{a}\)
Đặt \(t=\sqrt{a},t\ge0\)thì \(A=-t^2+t=-\left(t-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Suy ra Max A = 1/4 khi t = 0 => a = 1/4
Sửa đề :
a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)
b) \(A=4\)
\(\Leftrightarrow\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}=4\)
\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)
\(\Leftrightarrow x-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)
a)\(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)
b)\(\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
c)\(\sqrt{\left(4-\sqrt{17}\right)^2}=\left|4-\sqrt{17}\right|=\sqrt{17}-4\)
d)\(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}=2\sqrt{3}+\left|2-\sqrt{3}\right|=2\sqrt{3}+2-\sqrt{3}\)
Câu 1:
\(x^2-19=x^2-\left(\sqrt{19}^2\right)\left(x+\sqrt{19}\right)\)
Câu 2:
\(\sqrt{8t}.\sqrt{32t^3}=\sqrt{8t.32t^3}=\sqrt{\left(16.t^2\right)^2}=16.t^2\)
Câu 3 :
\(\sqrt{a^8\left(4-a\right)^2}=\sqrt{a.8}.\sqrt{\left(4-a\right)^2}=a^4\left|4-a\right|\)
( do \(a\le4\))
câu 1
\(x^2-19=\left(x-\sqrt{19}\right)\left(x+\sqrt{19}\right)\)
câu 2
\(\sqrt{8t}.\sqrt{32t^3}=\sqrt{8t.32t^3}=\sqrt{256t^4}=\sqrt{\left(16t^2\right)^2}=16t^2\)
câu 3
\(\sqrt{a^8\left(4-a\right)^2}=\sqrt{\left[a^4\left(4-a\right)\right]^2}=a^4\left(4-a\right)=4a^4-a^5\)
nếu mk sai thì bỏ qua nha <3
Đặt: \(a=\sqrt{2+x};b=\sqrt{2-x}\left(a,b\ge0\right)\)
\(\Rightarrow\hept{\begin{cases}a^2+b^2=4\\a^2-b^2=2x\end{cases}}\)
\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{4+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{4+ab}\)
\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a-b\right)\left(4+ab\right)}{4+ab}=\sqrt{2+ab}\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=\sqrt{4+2ab}\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(a^2+b^2+2ab\right)}\left(a-b\right)=\left(a+b\right)\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=a^2-b^2=2x\)
\(\Rightarrow A=x\sqrt{2}\)
\(\sqrt{\left(4+\sqrt2\right)^2}=\) \(\vert4+\sqrt2\vert\) \(=4+\sqrt2\)
\(\sqrt{\left(4+\sqrt2\right)^2}\)
\(=\left|4+\sqrt2\right|\)
\(=4+\sqrt2\)