\(A=\frac{\sqrt{z-5}}{z}+\frac{\sqrt{y-4}}{y}+\frac{\sqrt{x-3}}{x}=\frac{\sqrt{5\left(z-5\right)}}{\sqrt{5}z}+\frac{\sqrt{4\left(x-4\right)}}{2y}+\frac{\sqrt{3\left(x-3\right)}}{\sqrt{3}x}\)

Áp dụng BĐT Cosi ta có : \(A\le\frac{\frac{5+z-5}{2}}{\sqrt{5}z}+\frac{\frac{4+y-4}{2}}{2y}+\frac{\frac{3+x-3}{2}}{\sqrt{3}x}=\frac{\sqrt{5}}{10}+\frac{1}{4}+\frac{\sqrt{3}}{6}\)

Dấu "=" xảy ra \(\Leftrightarrow z=10;y=8;x=6\)

\(A=\frac{\sqrt{z-5}}{z}+\frac{\sqrt{y-4}}{y}+\frac{\sqrt{x-3}}{x}\)

Áp dụng bất đẳng thức Côsi:

\(z=z-5+5\ge2\sqrt{5.\left(z-5\right)}\)

\(\Rightarrow\frac{\sqrt{z-5}}{z}\le\frac{1}{2\sqrt{5}}\)

Dấu bằng xảy ra khi \(z-5=5\Leftrightarrow z=10\)

tương tự x, y.

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

đề thiếu điều kiện

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

Ta có: \(P=\frac{\sqrt{x}}{1+x+xy}+\frac{\sqrt{y}}{1+y+yz}+\frac{\sqrt{z}}{1+z+xz}\)

\(P=\frac{\sqrt{x}}{xy+x+1}+\frac{x\sqrt{y}}{x+xy+xyz}+\frac{xy\sqrt{z}}{xy+xyz+x^2yz}\)

\(P=\frac{\sqrt{x}}{xy+x+1}+\frac{x\sqrt{y}}{xy+x+1}+\frac{\sqrt{xy}.\sqrt{xyz}}{xy+x+1}\)

\(P=\frac{\sqrt{x}+x\sqrt{y}+\sqrt{xy}}{xy+x+1}\le\frac{\frac{x+1}{2}+\frac{x\left(y+1\right)}{2}+\frac{xy+1}{2}}{xy+x+1}\) (bđt cosi)

=> \(P\le\frac{x+1+xy+x+xy+1}{2\left(xy+x+1\right)}=\frac{2\left(xy+x+1\right)}{2\left(xy+x+1\right)}=1\)

Dấu "=" xảy ra<=> x =  y = z = 1

Vậy MaxP = 1 <=> x = y = z = 1

b, đk: \(x\ge1,y\ge2,z\ge3\)

\(=>B=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{y-2}=b\\\sqrt{z-3}=c\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}x=a^2+1\\y=b^2+1\\z=c^2+1\end{matrix}\right.\)\(=>a\ge0,b\ge0,c\ge0\)

B trở thành \(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\)

\(=\dfrac{a^{ }}{a^2+1}+\dfrac{a^2+1}{4}+\dfrac{b}{b^2+1}+\dfrac{b^2+1}{4}+\dfrac{c}{c^2+1}+\dfrac{c^2+1}{4}\)

\(-\left(\dfrac{a^2+b^2+c^2+3}{4}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}-\dfrac{a^2+b^2+c^2}{4}\)\(=0\)

dấu"=" xảy ra<=>\(a=0,b=0,c=0< =>x=1,y=2,z=3\)

Chắc bạn ghi nhầm đề, tìm GTLN mới đúng, chứ GTNN của các biểu thức này đều hiển nhiên bằng 0

\(A=\dfrac{3.\sqrt{x-9}}{15x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)

\(A_{max}=\dfrac{1}{30}\) khi \(x=18\)

\(B=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}=\dfrac{1.\sqrt{x-1}}{x}+\dfrac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}y}+\dfrac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}z}\)

\(B\le\dfrac{1+x-1}{2x}+\dfrac{2+y-2}{2\sqrt{2}y}+\dfrac{3+z-3}{2\sqrt{3}z}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;4;6\right)\)

đề bài là tìm gt lớn nhất nhé mọi người,tớ ghi nhầm

Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).

\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).

\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).

Ta có: 

\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)

\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).

Ta có:

\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).

\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).

\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).

\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).

Chứng minh tương tự, ta được:

\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).

Chứng minh tương tự, ta được:

\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).

\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).

\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)

\(\left(4\right)\).

Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).

\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)

(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).

\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(P\ge\frac{\sqrt{5}}{3}\).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).

Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).

đáp án

chia cả 2 vế của giả thiết cho xyz rồi đặt 1/x ; 1/y ; 1/z => a ; b ; c

đến đây thì tự làm tiếp đi