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a, ĐK: \(a\ge0;\) a khác 4
b,\(A=\frac{a+3\sqrt{a}+2+2a-4\sqrt{a}-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}+2\right)}=\frac{3a-6\sqrt{a}}{\left(\sqrt{a}+2\right)\cdot\left(\sqrt{a}-2\right)}=\frac{3\sqrt{a}}{\sqrt{a}+2}\)
c, để A= 2 KHI \(\frac{3\sqrt{a}}{\sqrt{a}+2}=2\)
<=>\(3\sqrt{a}=2\sqrt{a}+4\)
<=>\(\sqrt{a}=4\)
<=>a=16
tick nha
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
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
b.\(Q< 1\)
\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)
\(\Leftrightarrow4\sqrt{x}-8< 0\)
\(\Leftrightarrow0\le x< 4\)
Vay de Q<1 thi \(0\le0< 4\)
a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)
b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)
c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)
\(y\ne0\)
Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)
e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)
Vì y < 0 nên \(\left|y\right|=-y\)
Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)
f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)
- các cậu giúp mình với mai thứ 6 mình thi hk1 r huhu giúp mình với.........
- \(\sqrt{\frac{2ab^2}{162a}}=\sqrt{\frac{b^2}{81}}=\frac{|b|}{9}\)
- \(2y^2\sqrt{\frac{x^4}{4y^2}}=\frac{2y^2x^2}{-2y}=-yx^2\)
9 T I C H sai buồn
\(A=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\frac{1-x}{1-\sqrt{x}}..\)
nhờ vào năng lực rinegan tối hậu của ta , ta có thể dễ dàng nhìn thấy mẫu chung
\(x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=\sqrt{x}\left(\sqrt{x}-2\sqrt{xy}\right)+\left(\sqrt{x}-2\sqrt{y}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+1\right)\)
\(A=\frac{\sqrt{x^3}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}-\frac{2x\left(x-1\right)}{\left(\sqrt{x}-2\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}.\)
\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
\(A=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\sqrt{x}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\left(\sqrt{x}-2\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x}{\sqrt{y}}\)
b) thay y=625 vào ta được
\(\frac{x}{\sqrt{625}}=\frac{x}{25}< 0.2\Leftrightarrow x< 5\)
vậy \(0< x< 5\)
\(A=2\sqrt{27}-\sqrt{75}-\sqrt{\frac{4}{3}}\)\(=2\sqrt{9.3}-\sqrt{25.3}-\sqrt{\frac{4.3}{9}}\)\(=2.3\sqrt{3}-5\sqrt{3}-\frac{2}{3}\sqrt{3}\)\(=6\sqrt{3}-5\sqrt{3}-\frac{2}{3}\sqrt{3}\)\(=\frac{1}{3}\sqrt{3}\)\(=\frac{\sqrt{3}}{3}\)