Có \(4\left(\frac{9}{\sqrt{x-2}}+\sqrt{x-2}\right)\ge4.2\sqrt{\frac{9}{\sqrt{x-2}}\sqrt{x-2}}=24\)(Cô si)
\(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4}{\sqrt{y-1}}\sqrt{y-1}}=4\)
\(\Rightarrow\frac{4}{\sqrt{y-1}}+\sqrt{y-1}+4\left(\frac{9}{\sqrt{x-2}}+\sqrt{x-2}\right)\ge28\)
Dấu "=" xảy ra <=>\(\int^{9=x-2}_{4=y-1}\Leftrightarrow\int^{x=11}_{y=5}\)
giải pt : \(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}\)
bài 5
ĐK:\(x>2,y>1\)
\(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}..\)\(\Leftrightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)
Áp dụng AM-GM ta có:
\(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}\ge2\sqrt{\frac{144\sqrt{x-2}}{\sqrt{x-2}}}=24\)
\(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4\sqrt{y-1}}{\sqrt{y-1}}}=4.\)
\(\Rightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge28.\)
Dấu \(=\)xảy ra khi \(\frac{36}{\sqrt{x-2}}=4\sqrt{x-2}\Leftrightarrow x=11\left(n\right).\)
\(\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\Leftrightarrow y=5\left(n\right).\)
Vậy \(x=11,y=5\)
giải pt
1. \(x^2-5x+14=4\sqrt{x+1}\)
2. \(x^4+x^2+1=y^2\)với x, y nguyên
3. \(xy-2x+3y=21\)
4. \(x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\)
5. \(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}\)
6. \(2\left(x^2+2x+3\right)=5\sqrt{x^3+3x^2+3x+2}\)
giải phương trình \(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}\)
giúp mk vs
Chứng minh: \(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}\)
Tìm gia trị của x, y thỏa mãn phương trình :\(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4.\sqrt{x-2}-\sqrt{y-1}\)
Giải hệ pt:
1.\(\sqrt[4]{x}\left(\frac{1}{4}+\frac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=2\)
2.\(\sqrt[4]{y}\left(\frac{1}{4}-\frac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=1\)
giai p.t \(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}\)
Giải phương trình:
\(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}\) = \(28-4\sqrt{x-2}-\sqrt{y-1}\)