Các câu hỏi tương tự
+) Tìm min
\(E=\dfrac{1+\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{xy+yz+zx}\)
+) Tìm max và min
\(F=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\)
Trong đó a,b,c>0 và \(min\left\{a,b,c\right\}\ge\dfrac{1}{4}max\left\{a,b,c\right\}\)
cho x,y,z nguyen duong thoa man: \(\left\{{}\begin{matrix}\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\\\left|y-2x\right|\le\dfrac{1}{\sqrt{y}}\end{matrix}\right.\)
tim Max \(A=x^2+2y^2\)
GIẢI PT SAU:
\(\sqrt{3x-3}-\sqrt{5-x}=\sqrt{2x-4}\)
\(x^2-6x+9=4\sqrt{x^2-6x+6}\)
\(x^2-x+8-4\sqrt{x^2-x+4}=0\)
GIẢI CÁC PT SAU:
x2 - 6x + 9=\(4\sqrt{x^2-6x+6}\)
x2 - x + 8 - \(4\sqrt{x^2-x+4}=0\)
x2 + \(\sqrt{4x^2-12x+44}=3x+4\)
Cho 0≤x,y,z≤1
Tìm max \(D=\sqrt{\dfrac{x}{1+yz}}+\sqrt{\dfrac{y}{1+zx}}+\dfrac{z}{2+2xy}\)
a)\(\sqrt{x+5-4\sqrt{x+1}}+\sqrt{x+2-2\sqrt{x+1}}=1\) = 1
b) \(\sqrt{2x-2\sqrt{x-1}}-\sqrt{2\sqrt{2x+3}-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2\sqrt{2x-1}}}\)=4
GIÚP MÌNH VỚI. ĐANG CẦN GẤP
@Lia - Maths is fun !Let:a,b,cge0text{ }such:a+b+c3.Foundtext{ }maxtext{ }andtext{ }mintext{ }Asqrt{x+3}+sqrt{y+3}+sqrt{z+3} My solution !*Found maxUsing Bunhiacopxki we haveA^2leleft(a+3+b+3+c+3right)left(1+1+1right)...36Rightarrow Ale6left(Because:text{ }text{ }Age0text{ }sotext{ }Atext{ }canttext{ } 0text{ }right)A_{max}6text{ }Leftrightarrow abc1*Found minWe have extra inequality sqrt{x+z}+sqrt{y+z}gesqrt{z}+sqrt{x+y+z}left(x;y;zge0right)(1)Prove : l...
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@Lia - Maths is fun !
\(Let:a,b,c\ge0\text{ }such:a+b+c=3.Found\text{ }max\text{ }and\text{ }min\text{ }A=\sqrt{x+3}+\sqrt{y+3}+\sqrt{z+3}\)
My solution !
*Found max
Using Bunhiacopxki we have
\(A^2\le\left(a+3+b+3+c+3\right)\left(1+1+1\right)=...=36\)
\(\Rightarrow A\le6\left(Because\:\text{ }\text{ }A\ge0\text{ }so\text{ }A\text{ }can't\text{ }< 0\text{ }\right)\)
\(A_{max}=6\text{ }\Leftrightarrow a=b=c=1\)
*Found min
We have extra inequality \(\sqrt{x+z}+\sqrt{y+z}\ge\sqrt{z}+\sqrt{x+y+z}\left(x;y;z\ge0\right)\)(1)
Prove : \(\left(1\right)\Leftrightarrow x+y+2z+2\sqrt{\left(x+z\right)\left(y+z\right)}\ge z+x+y+z+2\sqrt{z\left(x+y+z\right)}\)
\(\Leftrightarrow\sqrt{xy+xz+yz+z^2}\ge\sqrt{xz+yz+z^2}\)
\(\Leftrightarrow xy+xz+yz+z^2\ge xz+yz+z^2\)
\(\Leftrightarrow xy\ge0\left(True!\right)\)
Using (1) we have
\(A=\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\ge\sqrt{3}+\sqrt{a+b+3}+\sqrt{c+3}\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{a+b+c}\)
\(=3\sqrt{3}\)
\(A_{min}=3\sqrt{3}\text{ }when\text{ }\hept{\begin{cases}a=b=\frac{3}{2}\\c=0\end{cases}}\)
(In here I using when because there are many other a,b,c such a = 0 ; b = c = 3/2)
The problem is done !
cho x,y,z>0 và x+y+z=3 Tìm Min của : \(P=\frac{x+y}{\sqrt{x^2+y^2+6z}}+\frac{y+z}{\sqrt{y^2+z^2+6x}}+\frac{z+x}{\sqrt{z^2+x^2+6y}}\)
\(\hept{\begin{cases}x^2-2y+10=\sqrt{x-2}+4\sqrt[3]{x+y}\\8\sqrt{x\left(y-2\right)}+4=8x+\left(y-x\right)^2\end{cases}}\)