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NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
\(y=\frac{1}{9+4\sqrt{5}}=\frac{1}{\left(\sqrt{5}+2\right)^2}\)
\(\Rightarrow N=\frac{1}{\left(\sqrt{5}-2\right)^2}-\frac{3}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{2}{9+4\sqrt{5}}\)
\(=\frac{1}{9-4\sqrt{5}}+\frac{2}{9+4\sqrt{5}}-3=\frac{9+4\sqrt{5}+18-8\sqrt{5}}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=24-4\sqrt{5}\)
\(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+x^2y^2-1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=2005^2-1\)
\(\Rightarrow S=\pm\sqrt{2005^2-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 6 2019
c/
Giả sử \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
\(\Leftrightarrow\sqrt[3]{3+\sqrt[3]{3}}-\sqrt[3]{3}< \sqrt[3]{3}-\sqrt[3]{3-\sqrt[3]{3}}\)
\(\Leftrightarrow\frac{\sqrt[3]{3}}{\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}}< \frac{\sqrt[3]{3}}{\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\) (1)
Ta có: \(\left\{{}\begin{matrix}\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}\\\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}>\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\end{matrix}\right.\)
Nên (1) đúng
Vậy BĐT ban đầu đúng
HS
1 tháng 8 2018
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
HS
1 tháng 8 2018
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
27 tháng 7 2022
Bài 1:
\(x=\dfrac{30-2\sqrt{45}}{4}=\dfrac{30-6\sqrt{5}}{4}< \dfrac{4\sqrt{17}}{4}=\sqrt{17}=y\)
LT
2 tháng 7 2019
a)\(\frac{3+\sqrt{3}}{1+\sqrt{3}}\)=\(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{1+\sqrt{3}}\)=\(\sqrt{3}\)
b)\(\frac{2\sqrt{3}-6}{\sqrt{8}-\sqrt{2}}\)
LT
2 tháng 7 2019
\(\frac{y-2\sqrt{y}}{\sqrt{y}-2}\)=\(\frac{\sqrt{y}\left(\sqrt{y}-2\right)}{\sqrt{y}-2}\)=\(\sqrt{y}\)
d) \(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}{\sqrt{x}-1}\)=\(\sqrt{x}\)+3
e)\(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)=\(\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)=\(\sqrt{y}\)-1
g)\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x+3}}\)
chúc bạn học tốt
BN
19 tháng 8 2017
a,Ta có :\(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\)
\(\Rightarrow x^3=4\left(\sqrt{5}+1\right)-4\left(\sqrt{5}-1\right)-3\sqrt[3]{4\left(\sqrt{5}-1\right).4\left(\sqrt{5}+1\right)}.\left(\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\right)\)\(\Rightarrow x^3=8-3\sqrt[3]{16\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}.x\)
\(\Rightarrow x^3=8-3\sqrt[3]{64}.x\Rightarrow x^3=8-12x\)\(\Rightarrow x^3-12x+8=0\)
Vậy \(x^3+12x-8=0\)
BN
19 tháng 8 2017
b,\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)(1)
Ta có :\(3=\left(x^2+3\right)-x^2=\left(\sqrt{x^2+3}-x\right)\left(\sqrt{x^2+3}+x\right)\)(2)
\(3=\left(y^2+3\right)-y^2=\left(\sqrt{y^2+3}-y\right)\left(\sqrt{y^2+3}+y\right)\) (3)
Từ (1) và (2) ta suy ra :\(y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)
Từ (1) và (3) ta suy ra :\(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)
Cộng 2 đẳng thức trên vế theo vế ta được :
\(x+y+\sqrt{x^2+3}+\sqrt{y^2+3}=\sqrt{x^2+3}+\sqrt{y^2+3}-x-y\)
\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
Vậy B=0
CW
24 tháng 6 2018
hình như...
b) \(x+y+z+8=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)
\(\Leftrightarrow x-3+y-3+z-3+17=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)
\(\Leftrightarrow\left(x-3-2\sqrt{x-3}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)+3=0\)
\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-3}-3\right)^2+3=0\) (vô nghiệm, VT >/3)
Kl: ptvn