đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)

ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)

\(\Leftrightarrow x^5-x^2\ge3x-3\)

cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)

\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)

\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)

áp dụng bunhia ta có:

\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)

cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)

đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)

\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)

\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)

\(\Rightarrow P\le1\)

\(\sqrt{x+\sqrt{x}}=y\)

ĐK:\(x\ge0,y\ge0\)

+)Nếu \(x=0\)thì suy ra \(y=0\).Do đó \(\left(x;y\right)=\left(0;0\right)\)

+)Nếu \(x>0\)thì:

\(\sqrt{x+\sqrt{x}}=y\Rightarrow x+\sqrt{x}=y^2\Rightarrow\sqrt{x}=y^2-x.\Rightarrow\sqrt{x}\)là số nguyên dương.

Đặt \(\sqrt{x}=t\)(t thuộc N*).Khi đó \(x=t^2\)và \(t^2+t=y^2.\)

Nhưng \(t^2< t^2+t=y^2< \left(t+1\right)^2.\)Điều này không xảy ra.

Vậy phương trình có 1 nghiệm duy nhất là \(\left(x;y\right)=\left(0;0\right).\)

Theo bdt bunhiacopxki

\(T^2=\left(x\sqrt{4-y^2}+y\sqrt{4-z^2}+z\sqrt{4-x^2}\right)^2\le\left(x^2+y^2+z^2\right)\left(12-x^2-y^2-z^2\right)\)

Đặt \(x^2+y^2+z^2=t\), bài toán ban đầu trở thành 

Cho t thuộc khoảng [0;12] tìm max của \(T^2=t\left(12-t\right)\)

Ta có\(\left(t-6\right)^2\ge0\Leftrightarrow t^2-12t+36\ge0\Rightarrow t^2-12t\ge-36\Rightarrow-t^2+12t\le36\)

\(\Rightarrow t\left(12-t\right)\le36\Leftrightarrow T^2\le36\Rightarrow T\le6\) Vậy max=6, dấu bằng xảy ra khi \(x=y=z=\sqrt{2}\)

ta có bđt cần chứng minh 

\(\frac{\sqrt{xy+z}+\sqrt{2x^2+2y^2}}{1+\sqrt{xy}}\ge1\Leftrightarrow\sqrt{xy+z}+\sqrt{2\left(x^2+y^2\right)}\ge1+\sqrt{xy}\)

Áp dụng bđt bu nhi ta có 

\(\sqrt{2\left(x^2+y^2\right)}\ge x+y\) (1)

mà x+y+z=1\(\Rightarrow xy+z=xy+z\left(x+y+z\right)=\left(z+x\right)\left(z+y\right)\)

áp dụng bu nhi a ta có \(\sqrt{\left(z+x\right)\left(z+y\right)}\ge z+\sqrt{xy}\) (2)

từ (1) và (2) => \(\sqrt{xy+z}+\sqrt{2x^2+2y^2}\ge x+y+z+\sqrt{xy}=1+\sqrt{xy}\)

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 \(x=11,y=5\)

\(\left(\sqrt{8-\sqrt{X-3}}-\sqrt{5-\sqrt{X-3}}\right)^2=5^2\)=\(5^2\)

\(\Leftrightarrow-2\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=25-3=22\)

\(\Leftrightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=11\)

Do \(\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\ge0\Rightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\le0\)

\(\Rightarrow\)PT vô nghiệm

Tìm giá trị của x,y,z, biết:

1) x + y + z + 8= 2\(\sqrt{x-1}\)+ 4\(\sqrt{y-2}\)+ 6\(\sqrt{z-3}\)
2) x + y + z= 2\(\sqrt{x}\)+ 2\(\sqrt{y-3}\)+ 2\(\sqrt{z}\)
3) x + y + 4= 2\(\sqrt{x}\)+ 4\(\sqrt{y-1}\)
4) \(\sqrt{x+3}\)+ \(\sqrt{y-2}+\sqrt{z-3}\)= \(\frac{1}{2}\)(x + y + z + 1)
5)\(\sqrt{x-2}+\sqrt{y+2009}+\sqrt{z-2010}=\)\(\frac{1}{2}\)(x + y + z)

\(\frac{1}{2}\)

Tim x, y, z

1/ \(\sqrt{x-2}+\sqrt{y-2008}+\sqrt{z-2009}=\dfrac{1}{2}\left(x+y+z\right)\)

2/ \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{x-5}\)

3/ Tinh T = \(x^2+y^2+z^2-7\) biet x-y-z = \(2\sqrt{x-34}+4\sqrt{y-21}+6\sqrt{z-4}+45\)

4/ \(2x^2+9y^2-6xy-12y-6x+29=0\)

5/\(4x^2+3y-4x+4xy-10y+9=0\)