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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: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
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) \(\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}}-\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}-1}}=\dfrac{\sqrt{2}\left(\sqrt{\sqrt{2}-1}\right)}{\left(\sqrt{\sqrt{2}+1}\right)\left(\sqrt{\sqrt{2}-1}\right)}-\dfrac{\sqrt{2}\left(\sqrt{\sqrt{2}+1}\right)}{\left(\sqrt{\sqrt{2}+1}\right)\left(\sqrt{\sqrt{2}-1}\right)}=\dfrac{\sqrt{2}\left(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\right)}{\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\dfrac{\sqrt{2}\left(\left(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\right)\right)}{\sqrt{2-1}}=\sqrt{2}.\left(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\right)\)(1)
Đặt A=\(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\Leftrightarrow A^2=\sqrt{2}-1+\sqrt{2}+1-2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}-2\sqrt{1}=2\sqrt{2}-2\Leftrightarrow A=\pm\sqrt{2\sqrt{2}-2}\)
Ta có \(\sqrt{\sqrt{2}-1}< \sqrt{\sqrt{2}+1}\Leftrightarrow\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}< 0\Leftrightarrow A< 0\)
Vậy A=\(-\sqrt{2\sqrt{2}-2}\)
(1)\(=\sqrt{2}.\left(-\sqrt{2\sqrt{2}-2}\right)=-\sqrt{4\sqrt{2}-4}\)
b) \(\sqrt{4-2\sqrt{3}}+\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{27}=\sqrt{3-2.\sqrt{3}.1+1}+\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{9.3}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\dfrac{4+2\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}}-3\sqrt{3}=\left|\sqrt{3}-1\right|+\sqrt{4+2\sqrt{3}}-3\sqrt{3}=\sqrt{3}-1-3\sqrt{3}+\sqrt{3+2\sqrt{3}+1}=-2\sqrt{3}-1+\sqrt{\left(\sqrt{3}+1\right)^2}=-2\sqrt{3}-1+\sqrt{3}+1=-\sqrt{3}\)
c) \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-3\sqrt{x}-2\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left[3\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3\sqrt{x}-2\right)}{\sqrt{x}+3}=\dfrac{2-3\sqrt{x}}{\sqrt{x}+3}\)
Áp dụng BĐT cô-si,ta có:
\(\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)>=2 căn bậc 4 của x.y
Vậy A>= -căn x-căn y - căn 4 của x.y=-(\(\sqrt{x}-\sqrt{y}\) )^2
Mà (\(\left(\sqrt{x}-\sqrt{y}\right)^2>=0\)
Suy ra A<=0
dấu = xảy ra khi và chỉ khi x=y
Sử dụng BĐT Cauchy-Schwarz, ta có:
\(VP=\sqrt{x}.\sqrt{1-\dfrac{1}{x}}+\sqrt{y}.\sqrt{1-\dfrac{1}{y}}+\sqrt{z}.\sqrt{1-\dfrac{1}{z}}\le\sqrt{\left(x+y+z\right)\left(1-\dfrac{1}{x}+1-\dfrac{1}{y}+1-\dfrac{1}{z}\right)}=\sqrt{x+y+z}=VT\)
Chứng minh hoàn tất.