Các câu hỏi tương tự
1,cmr frac{2sqrt{mn}}{sqrt{n}+sqrt{n}+sqrt{m+n}}sqrt{m}+sqrt{n}-sqrt{m+n}1,rút gọn a, 3sqrt{27a}+2sqrt{frac{a}{3}}+asqrt{frac{4}{3a}}b,x^2sqrt{frac{12y}{x}}-xysqrt{frac{x}{3y}}c,frac{x}{left(sqrt{x}-sqrt{y}right)left(sqrt{x}-sqrt{z}right)}+frac{y}{left(sqrt{y}-sqrt{z}right)left(sqrt{y}-sqrt{x}right)}+frac{z}{left(sqrt{z}-sqrt{x}right)left(sqrt{z}-sqrt{y}right)}
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1,cmr
\(\frac{2\sqrt{mn}}{\sqrt{n}+\sqrt{n}+\sqrt{m+n}}\)=\(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)
1,rút gọn
a, 3\(\sqrt{27a}+2\sqrt{\frac{a}{3}}+a\sqrt{\frac{4}{3a}}\)
b,\(x^2\sqrt{\frac{12y}{x}}-xy\sqrt{\frac{x}{3y}}\)
c,\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)
CHO a,b,c0 thỏa mãn: a^2b^2+b^2c^2+c^2a^2ge a^2+b^2+c^2CMR: frac{a^2b^2}{c^3left(a^2+b^2right)}+frac{b^2c^2}{a^3left(b^2+c^2right)}+frac{c^2a^2}{b^3left(a^2+c^2right)}gefrac{sqrt{3}}{2}ĐẶT Afrac{a^2b^2}{c^3left(a^2+b^2right)}+frac{b^2c^2}{a^3left(b^2+c^2right)}+frac{c^2a^2}{b^3left(c^2+a^2right)}ĐẶT:frac{1}{a}x,frac{1}{y}b,frac{1}{z}cRightarrow x^2+y^2+z^2ge1Rightarrow Afrac{x^3}{y^2+z^2}+frac{y^3}{z^2+x^2}+frac{z^3}{z^2+y^2}TA CÓ:xleft(y^2+z^2right)frac{1}{sqrt{2}}sqrt{2x^2left(y^2+z^2right)lef...
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CHO a,b,c>0 thỏa mãn: \(a^2b^2+b^2c^2+c^2a^2\ge a^2+b^2+c^2\)
CMR: \(\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(a^2+c^2\right)}\ge\frac{\sqrt{3}}{2}\)
ĐẶT \(A=\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(c^2+a^2\right)}\)
ĐẶT:\(\frac{1}{a}=x,\frac{1}{y}=b,\frac{1}{z}=c\)
\(\Rightarrow x^2+y^2+z^2\ge1\)
\(\Rightarrow A=\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{z^2+y^2}\)
TA CÓ:
\(x\left(y^2+z^2\right)=\frac{1}{\sqrt{2}}\sqrt{2x^2\left(y^2+z^2\right)\left(y^2+z^2\right)}\le\frac{1}{\sqrt{2}}\sqrt{\frac{\left(2x^2+2y^2+2z^2\right)^3}{27}}=\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)TƯƠNG TỰ:
\(y\left(x^2+z^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2},z\left(x^2+y^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)LẠI CÓ:
\(A=\frac{x^3}{y^2+z^2}+\frac{y^3}{x^2+z^2}+\frac{z^3}{x^2+y^2}=\frac{x^4}{x\left(y^2+z^2\right)}+\frac{y^4}{y\left(x^2+z^2\right)}+\frac{z^4}{z\left(x^2+y^2\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)}\ge\frac{1}{3.\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}}
\)\(\ge\frac{\sqrt{3}}{2}\sqrt{x^2+y^2+z^2}\ge\frac{\sqrt{3}}{2}\)
DẤU BẰNG XẢY RA\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\Rightarrow DPCM\)
đặt Afrac{sqrt{yz}}{x+3sqrt{yz}}+frac{sqrt{zx}}{y+3sqrt{zx}}+frac{sqrt{xy}}{z+3sqrt{xy}}Rightarrow1-3Afrac{x}{x+3sqrt{yz}}+frac{y}{y+3sqrt{zx}}+frac{z}{z+3sqrt{xy}}gefrac{x}{x+frac{3}{2}left(y+zright)}+frac{y}{y+frac{3}{2}left(z+xright)}+frac{z}{z+frac{3}{2}left(x+yright)}frac{2x}{2x+3left(y+zright)}+frac{2y}{2y+3left(z+xright)}+frac{2z}{2z+3left(x+yright)}frac{2x^2}{2x^2+3xy+3xz}+frac{2y^2}{2y^2+3yz+3xy}+frac{2z^2}{2z^2+3zx+3yz}gefrac{2left(x+y+zright)^2}{2left(x^2+y^2+z^2right)+6left(xy+yz+zxr...
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đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
Cho C = \(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]\left[\frac{x\sqrt{x}+y\sqrt{z}+x\sqrt{y}+y\sqrt{y}}{\sqrt{x^3y}+\sqrt{xy^3}}\right]...\)
a) Rút gọn C
b) Tìm x,y biết xy= \(\frac{1}{16}\)và C = 5
chị QAta có đề bài frac{x^2}{y}-2x+y+frac{y^2}{z}-2y+z+frac{z^2}{x}-2z+x+left(x+y+zright)-left(x-yright)^2-left(y-zright)^2-left(z-xright)^2frac{left(x-yright)^2}{y}-left(x-yright)^2+...+left(x+y+zright)left(x-yright)^2left(frac{1}{y}-1right)+....+left(x+y+zright)mà sqrt{x}+sqrt{y}+sqrt{z}1Rightarrow x,y,zinleft[0;1right] frac{1}{y}-y0 Age x+y+zgefrac{left(sqrt{x}+sqrt{y}+sqrt{z}right)^2}{3}frac{1}{3}
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chị QA
ta có đề bài <=>
\(\frac{x^2}{y}-2x+y+\frac{y^2}{z}-2y+z+\frac{z^2}{x}-2z+x+\left(x+y+z\right)-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2\)
=\(\frac{\left(x-y\right)^2}{y}-\left(x-y\right)^2+...+\left(x+y+z\right)\)
=\(\left(x-y\right)^2\left(\frac{1}{y}-1\right)+....+\left(x+y+z\right)\)
mà \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\Rightarrow x,y,z\in\left[0;1\right]\)
=> \(\frac{1}{y}-y>0\)
=> \(A\ge x+y+z\ge\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=\frac{1}{3}\)
Tìm max
\(A=3\sqrt{2x-1}+x\sqrt{5-4x^2}\left(\frac{1}{2}\le x\le\frac{\sqrt{5}}{2}\right)\)
\(B=\frac{xyz\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}\left(x,y,z>0\right)\)
\(\sqrt{x^2+xy+y^2}=\sqrt{\left(x+y\right)^2-xy}\ge\sqrt{\left(x+y\right)^2-\frac{1}{4}\left(x+y\right)^2}=\frac{x+y}{2}.\sqrt{3}\)
cmtt=>\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)=3\)
Cho 3 số dương a y z thỏa mãn xyz=1 ,tìm GTNN của
P= \(\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(y+x\right)}{x\sqrt{x}+2y\sqrt{y}}\)
Cho 3 số dương x,y,z thỏa mãn: xy+yz+zx=1. Chứng minh rằng:
\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+z^2\right)\left(1+y^2\right)}{1+z^2}}=2\)