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a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
1) ta có : \(P=\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow P=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(\Leftrightarrow P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)=x-y\)
2) ta có : \(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(B=\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}-\dfrac{4}{2\sqrt{6}}+\dfrac{\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{3}\sqrt{4+2\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}-4}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}+1\right)^2-4}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}:\left(\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+3\right)}{2\sqrt{6}}\right)\)
\(B=\dfrac{\sqrt{3}+1}{2\sqrt{2}}.\dfrac{2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\sqrt{3}}=\dfrac{1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{6}-\sqrt{2}}\)
\(\Leftrightarrow B=\dfrac{\left(\sqrt{6}+\sqrt{2}\right)}{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
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) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
\(A=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
\(\text{a) }\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\\ =\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\\ =\sqrt{xy}\)
\(\text{b) }\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(\text{c) }\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}\\ =\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}\\ =\dfrac{\sqrt{y}-1}{x-1}\)
a)\(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{x}\sqrt{y}+y\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}+y\)
\(=x+\sqrt{xy}+y-x+2\sqrt{xy}+y\)
\(=3\sqrt{xy}+2y\)
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
Đề câu c co bị sai ko vậy bạn? (y - 2\(\sqrt{x}\) +1)
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)