Tính: S=\(5x^2+6xy+5y^2\) với \(x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) và \(y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
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MH
7 tháng 7 2017
có: ở x ta nhân cả tử và mẫu với\(\sqrt{3}+\sqrt{2}\) ta được \(x=2\left(\sqrt{3}+\sqrt{2}\right)=\sqrt{12}+\sqrt{8}\)
ở y ta nhân cả tử và mẫu với \(\sqrt{3}-\sqrt{2}\)ta được
\(y=2\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{12}-\sqrt{8}\)
thay x và y vào A ta dc :
\(5\left(\sqrt{12}+\sqrt{8}\right)^2+6\left(\sqrt{12}+\sqrt{8}\right)\left(\sqrt{12}-\sqrt{8}\right)+5\left(\sqrt{12}-\sqrt{8}\right)=5\left(24+16\right)+24=224\)
mình cx ko chắc lắm ddaaau nha
AH
Akai Haruma
Giáo viên
16 tháng 3 2018
Lời giải:
Ta có: \(5x^2+6xy+5y^2=3(x^2+y^2+2xy)+2(x^2+y^2)\)
\(=3(x+y)^2+2(x^2+y^2)\geq 3(x+y)^2+(x+y)^2\) (theo BĐT AM-GM)
\(\Leftrightarrow 5x^2+6xy+5y^2\geq 4(x+y)^2\Rightarrow \sqrt{5x^2+6xy+5y^2}\geq 2(x+y)\)
Thực hiện tương tự với những biểu thức còn lại suy ra:
\(P\geq \frac{2(x+y)}{x+y+2z}+\frac{2(y+z)}{y+z+2x}+\frac{2(z+x)}{z+x+2y}\)
\(P\geq 2\left(\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}\right)=2\left(\frac{(x+y)^2}{(x+y+2z)(x+y)}+\frac{(y+z)^2}{(y+z+2x)(y+z)}+\frac{(z+x)^2}{(z+x+2y)(z+x)}\right)\)
Áp dụng BĐT Cauchy-Schwarz:
\(P\geq 2.\frac{(x+y+y+z+z+x)^2}{(x+y+2z)(x+y)+(y+z+2x)(y+z)+(z+x+2y)(z+x)}\)
\(\Leftrightarrow P\geq 2. \frac{4(x+y+z)^2}{2(x+y+z)^2+2(xy+yz+xz)}=\frac{4(x+y+z)^2}{(x+y+z)^2+xy+yz+xz}\)
\(\geq \frac{4(x+y+z)^2}{(x+y+z)^2+\frac{(x+y+z)^2}{3}}=3\) (theo AM-GM \(xy+yz+xz\leq \frac{(x+y+z)^2}{3}\))
Vậy \(P\geq 3\Leftrightarrow P_{\min}=3\)
Dấu bằng xảy ra khi \(x=y=z\)
14 tháng 1 2018
Ta có: \(5x^2+6xy+5y^2=4\left(x+y\right)^2+\left(x-y\right)^2\ge4\left(x+y\right)^2\)
tương tự: \(5y^2+6yz+5z^2\ge4\left(y+z\right)^2\) ;\(5z^2+6xz+5z^2\ge4\left(x+z\right)^2\)
\(\Rightarrow P\ge\dfrac{2\left(x+y\right)}{x+y+2z}+\dfrac{2\left(y+z\right)}{y+z+2x}+\dfrac{2\left(x+z\right)}{x+z+2y}\)
\(\Leftrightarrow\dfrac{P}{2}\ge\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{y+z+2x}+\dfrac{x+z}{x+z+2y}\)
\(\Leftrightarrow\dfrac{P}{2}\ge\dfrac{x+y}{\left(x+z\right)+\left(y+z\right)}+\dfrac{y+z}{\left(x+y\right)+\left(x+z\right)}+\dfrac{x+z}{\left(x+y\right)+\left(y+z\right)}\)Theo BDT Nesbit
\(\dfrac{x+y}{\left(x+z\right)+\left(y+z\right)}+\dfrac{y+z}{\left(x+y\right)+\left(x+z\right)}+\dfrac{x+z}{\left(x+y\right)+\left(y+z\right)}\ge\dfrac{3}{2}\)
Vậy \(\dfrac{P}{2}\ge\dfrac{3}{2}\Leftrightarrow P\ge3\)
Min P = 3 khi x = y = z
6 tháng 7 2018
\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3+1+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3+1-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\ =\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\\ =\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{\left(2\sqrt{2}-\sqrt{6}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\dfrac{2\sqrt{6}+3\sqrt{2}-2\sqrt{2}-\sqrt{6}+2\sqrt{6}-3\sqrt{2}+2\sqrt{2}-\sqrt{6}}{2\sqrt{3}}\\ =\dfrac{4\sqrt{6}-2\sqrt{6}}{2\sqrt{3}}=\dfrac{\sqrt{6}\left(4-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}}{2\sqrt{3}}=\sqrt{2}\)
\(y=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}+1}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}-1}\\ =\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{5}-1}\\ =\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\dfrac{\left(3\sqrt{2}-\sqrt{10}\right)\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\\ =\dfrac{3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}-3\sqrt{10}+5\sqrt{2}-3\sqrt{2}+\sqrt{10}}{4}\\ =\dfrac{4\sqrt{2}}{4}=\sqrt{2}\)
Vậy \(x=y\)
\(x=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\left(\sqrt{3}+\sqrt{2}\right)^2=5+2\sqrt{6}\)
\(y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\left(\sqrt{3}-\sqrt{2}\right)^2=5-2\sqrt{6}\)
\(S=5\left(5+2\sqrt{6}\right)^2+6\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)+5\left(5-2\sqrt{6}\right)^2\)
\(=5\left(49+20\sqrt{6}\right)+6+5\left(49-20\sqrt{6}\right)\)
\(=245+245+6=496\)