Chứng minh rằng : (1+2x).(1+\(\frac{y}{2x}\)).\(\left(1+\frac{4}{\sqrt{y}}\right)^2\)≥81 √x,y>0
Những câu hỏi liên quan
cho x,y>0 và 2x>y Chứng minh rằng \(\left(\frac{1}{x}+2\right)^2.\left(\frac{2}{y}-\frac{1}{x}\right).\frac{2y-1}{y}< =\frac{81}{8}\)
CHO \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
Chứng minh rằng :\(\left(x^2y^2+y^2z^2+z^2x^2\right)^2=2\left(x^4y^4+y^4z^4+z^4x^4\right)\)
GIÚP MÌNH VỚI
Ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{yz+zx+xy}{xyz}=0\) (Quy đồng)
\(\Rightarrow yz+zx+xy=0\)
Vì:
\(\left(x^2y^2+y^2z^2+z^2x^2\right)^2=0\)
\(2\left(x^4y^{ }^4+y^4z^4+z^4x^4\right)=0\)
Nên.....(tự kết luận nha)
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giải chi tiết ( vì sao ) đoạn dưới đây = 0 hộ mk vs :
vì \(\left(x^2y^2+y^2z^2+z^2x^2\right)^2=0\)
\(2\left(x^4y^4+y^4z^4+z^4x^4\right)=0\)
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-Ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow xy+yz+zx=0\)
Đặt \(xy=a,yz=b,zx=c\) thì bài toán thành
Cho \(a+b+c=0\)chứng minh \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)
Ta có:
\(\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)
\(=2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\)
\(=c^2\left(a+b\right)^2+c^2\left(a-b\right)^2-\left(a^2-b^2\right)^2-c^4\)
\(=c^2\left[\left(a+b\right)^2-c^2\right]+\left(a-b\right)^2\left[c^2-\left(a+b\right)^2\right]\)
\(=c^2\left(a+b+c\right)\left(a+b-c\right)+\left(a-b\right)^2\left(a+b+c\right)\left(c-a-b\right)\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left[c^2-\left(a-b\right)^2\right]=0\)
Vậy \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)
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Rút gon
A = \(\left(\sqrt{6x^2-12xy^2+6y^3}+\sqrt{24x^2y}\right):\sqrt{6y}\)
B = \(\frac{\sqrt{343xy^3\left(x-y\right)^2}}{\sqrt{28xy}}\) với x, y>0 , x<y
C= \(\sqrt{\frac{m}{1-2x+x^2}}:\frac{\sqrt{81}}{4m^3\left(x^2-2x+1\right)}\) với m>0 , m khác 1
\(A=\left(\sqrt{6\left(x^2-2xy^2+y^3\right)}+\sqrt{6.4x^2y}\right).\frac{1}{\sqrt{6y}}\)
\(=\left(\sqrt{6\left(x^2-xy^2+y^3\right)}+2x\sqrt{6y}\right).\frac{1}{\sqrt{6y}}\)
\(=\left[\sqrt{6}\left(\sqrt{x^2-xy^2+y^3}+2x\sqrt{y}\right)\right].\frac{1}{\sqrt{6y}}=\sqrt{6}\left(\sqrt{x^2-xy^2+y^3}-2x\sqrt{y}\right).\frac{1}{\sqrt{6}\sqrt{y}}\)
\(=\frac{x^2-xy^2+y^3}{\sqrt{y}}-\frac{2x\sqrt{y}}{\sqrt{y}}=\frac{x^2-xy^2+y^3}{\sqrt{y}}-2x\)
mik chỉ lm đến đây đc thui
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\(B=\frac{7y\left(y-x\right)\sqrt{7xy}}{2\sqrt{7xy}}=7y^2-7x\)
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\(C=\frac{\sqrt{m}}{\sqrt{\left(x-1\right)^2}}.\frac{4m^3\left(x-1\right)^2}{9}=\frac{\sqrt{m}}{\left(x-1\right)}.\frac{4m^3\left(x-1\right)^2}{9}=\frac{4m^3\sqrt{m}\left(x-1\right)}{9}\)
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Giải hpt sau:
a) left{{}begin{matrix}sqrt{5}x+left(1-sqrt{3}right)y1left(1-sqrt{3}right)x+sqrt{5}y1end{matrix}right.
b)left{{}begin{matrix}frac{3x}{x+1}-frac{2y}{y+4}4frac{2x}{x+1}-frac{5y}{y+4}5end{matrix}right.
c) left{{}begin{matrix}3x-2left|yright|92x+3left|yright|1end{matrix}right.
d) left{{}begin{matrix}frac{2}{2x-y}+frac{3}{x-2y}frac{1}{2}frac{2}{2x-y}-frac{1}{x-2y}frac{1}{18}end{matrix}right.
Đọc tiếp
Giải hpt sau:
a) \(\left\{{}\begin{matrix}\sqrt{5}x+\left(1-\sqrt{3}\right)y=1\\\left(1-\sqrt{3}\right)x+\sqrt{5}y=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{3x}{x+1}-\frac{2y}{y+4}=4\\\frac{2x}{x+1}-\frac{5y}{y+4}=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{2}{2x-y}+\frac{3}{x-2y}=\frac{1}{2}\\\frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18}\end{matrix}\right.\)
Chứng minh rằng:
Với x>1; y>0 thì \(\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)\)
Với x;y;z> 0 thoả mãn hệ thức \(x+y+z=18\sqrt{2}\)
Chứng minh rằng : \(\frac{1}{\sqrt{x\left(y+z\right)}}+\frac{1}{\sqrt{y\left(z+x\right)}}+\frac{1}{\sqrt{z\left(x+y\right)}}\ge\frac{1}{4}\)
Côsi: \(\sqrt{x\left(y+z\right)}=\frac{1}{2\sqrt{2}}.2.\sqrt{2x}.\sqrt{y+z}\le\frac{1}{2\sqrt{2}}\left(2x+y+z\right)\)
\(\Rightarrow\frac{1}{\sqrt{x\left(y+z\right)}}\ge\frac{2\sqrt{2}}{2x+y+z}\)
Tương tự các cái kia.
\(\Rightarrow VT\ge2\sqrt{2}\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)
\(\ge2\sqrt{2}.\frac{9}{2x+y+z+2y+z+x+2z+x+y}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)
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\(\sum\frac{1}{\sqrt{x\left(y+z\right)}}=\sum\frac{\sqrt{2}}{\sqrt{2x}.\sqrt{y+z}}\ge\sum\frac{2\sqrt{2}}{2x+y+z}\ge2\sqrt{2}.\frac{9}{\sum\left(2x+y+z\right)}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)
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Cho \(\hept{\begin{cases}x,y,z>0\\xy+yz+zx=1\end{cases}}\). Chứng minh rằng:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge3+\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x^2}}+\sqrt{\frac{\left(y+z\right)\left(y+x\right)}{y^2}}+\sqrt{\frac{\left(z+x\right)\left(z+y\right)}{z^2}}\)
1111111111111111111
\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
Bài 1 cho x,y,z>2014 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{1007}\)
chứng minh rằng \(\sqrt{x+y+z}\ge\sqrt{x-2014}+\sqrt{y-2014}+\sqrt{z-2014}\)
Bài 2
cho a,b,c>0. chứng minh rằng
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}\ge\frac{4}{ab+bc+ca}\)
Bài 2 : đã cm bên kia
Bài 1: :|
we had điều này:
\(2=\frac{2014}{x}+\frac{2014}{y}+\frac{2014}{z}\)
\(\Leftrightarrow\frac{x-2014}{x}+\frac{y-2014}{y}+\frac{z-204}{z}=1\)
Xòng! bunyakovsky
P/s : Bệnh lười kinh niên tái phát nên ít khi ol sorry :<
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Cho x,y>0 Chứng minh rằng:\(\left(x+1\right)\left(1+\frac{y}{x}\right)\left(1+\frac{9}{\sqrt{y}}\right)^2\ge256\)
\(\left(1+x\right)\left(1+\frac{y}{x}\right)\ge\left(1+\sqrt{\frac{x.y}{x}}\right)^2=\left(1+\sqrt{y}\right)^2\)
\(\Rightarrow VT\ge\left[\left(1+\sqrt{y}\right)\left(1+\frac{9}{\sqrt{y}}\right)\right]^2\ge\left(1+3\right)^4=256\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y=9\\x=3\end{matrix}\right.\)