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Theo đề: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow-\left(x+y\right)=z\)
\(\Leftrightarrow-\left(x+y\right)^5=z^5\)
\(x^2+y^2+z^2=1\)
\(\Rightarrow x^2+y^2=1-z^2\)
\(\Rightarrow\left(x+y\right)^2-2xy=1-z^2\)
\(\Rightarrow\left(x+y\right)^2=1-z^2+2xy\)
\(\Rightarrow\left(-z\right)^2=1-z^2+2xy\)
\(\Leftrightarrow xy=\frac{2z^2-1}{2}\)
Nên ta có:
\(VT=x^5+y^5+z^5=x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)
\(=x^5+y^5-x^5-5x^4y-10x^3y^2-10x^2y^3-5xy^4-y^5\)
\(=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)
\(=-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\)
\(=-5xy\left(x+y\right)\left(x^2-xy+y^2\right)-10x^2y^2\left(x+y\right)\)
\(=-5xy\left(x+y\right)\left(x^2-xy+y^2+2xy\right)\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
\(=-5.\frac{2z^2-1}{2}.\left(-z\right).\left(1-z^2+\frac{2z^2-1}{2}\right)\)
\(=\frac{5z\left(2z^2-z\right)}{4}=\frac{5}{4}z\left(2x^2-1\right)=\frac{5}{4}\left(2z^3-z\right)=VP\)
=> đpcm
Cộng vế theo vế
=> \(x^2+x+y^2+y+z^2+z=x^2+y^2+z^2\)
=> \(x+y+z=0\)=> A = 0
\(x=\left(y^2-x^2\right)=\left(y-x\right)\left(y+x\right)=\left(y-x\right).\left(-z\right)=\left(x-y\right).z\)
\(y=\left(z-y\right)\left(z+y\right)=\left(z-y\right).-x=x\left(y-z\right)\)
\(z=y\left(z-x\right)\)
=> \(xyz=\left(x-y\right)\left(y-z\right)\left(z-x\right).xyz\)
=> B = 1
Bài 2. a/ \(1\le a,b,c\le3\) \(\Rightarrow\left(a-1\right).\left(a-3\right)\le0\) , \(\left(b-1\right)\left(b-3\right)\le0\), \(\left(c-1\right).\left(c-3\right)\le0\)
Cộng theo vế : \(a^2+b^2+c^2\le4a+4b+4c-9\)
\(\Rightarrow a+b+c\ge\frac{a^2+b^2+c^2+9}{4}=7\)
Vậy min E = 7 tại chẳng hạn, x = y = 3, z = 1
b/ Ta có : \(x+2y+z=\left(x+y\right)+\left(y+z\right)\ge2\sqrt{\left(x+y\right)\left(y+z\right)}\)
Tương tự : \(y+2z+x\ge2\sqrt{\left(y+z\right)\left(z+x\right)}\) , \(z+2y+x\ge2\sqrt{\left(z+y\right)\left(y+x\right)}\)
Nhân theo vế : \(\left(x+2y+z\right)\left(y+2z+x\right)\left(z+2y+x\right)\ge8\left(x+y\right)\left(y+z\right)\left(z+x\right)\) hay
\(\left(x+2y+z\right)\left(y+2z+x\right)\left(z+2y+x\right)\ge64\)
=2/(xy+yz+zx)+2/(x^2+y^2+z^2)+1/xy+yz+zx
>=2(4/(x+y+z)^2)+1/(1/3)>=8+3=11(hình như sai đề nhưng cách làm là đúng rồi)
=2/(xy+yz+zx)+2/(x^2+y^2+z^2)+1/xy+yz+zx
>=2(4/(x+y+z)^2)+1/(1/3)>=8+3=11(hình như sai đề nhưng cách làm là đúng rồi)
1/ Ta cần c/m: \(3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
Tức là \(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)
Ta có đpcm.
\(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=12\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow\)\(4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\le\frac{49}{16}\)
\(\Leftrightarrow\)\(\left[2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\right]^2\le\frac{49}{16}\)
\(\Leftrightarrow\)\(\frac{-7}{4}\le2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\le\frac{7}{4}\)
\(\Leftrightarrow\)\(\frac{-3}{4}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le1\)
Có : \(\frac{1}{4a+b+c}+\frac{1}{a+4b+c}+\frac{1}{a+b+4c}\le\frac{1}{36}\left(\frac{6}{a}+\frac{6}{b}+\frac{6}{c}\right)\le\frac{1}{6}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=3\)
...
voi x,y,z>0 ta co
ap dung bdt co si ta co
\(T>=3\sqrt[3]{\sqrt{\left(\frac{x^2+1}{x^2}+\frac{1}{y^2}\right)\left(\frac{y^2+1}{y^2}+\frac{1}{z^2}\right)\left(\frac{z^2+1}{z^2}+\frac{1}{x^2}\right)}}\)
=\(3\sqrt[3]{\sqrt{\left(1+\frac{1}{x^2}+\frac{1}{y^2}\right)\left(1+\frac{1}{y^2}+\frac{1}{z^2}\right)\left(1+\frac{1}{z^2}+\frac{1}{x^2}\right)}}\)
>=\(3\sqrt[3]{\sqrt{3\sqrt[3]{\frac{1}{x^2y^2}}.3\sqrt[3]{\frac{1}{y^2z^2}}.3\sqrt[3]{\frac{1}{x^2z^2}}}}=3\sqrt[3]{\sqrt{27\sqrt[3]{\frac{1}{\left(xyz\right)^4}}}}\)
=\(3\sqrt[3]{\sqrt{27.\frac{1}{xyz}.\sqrt[3]{\frac{1}{xyz}}}}=3\sqrt{3}.\sqrt[9]{\frac{1}{\left(xyz\right)^2}}\)
ap dung bdt co si ta co
\(x+y+z>=3\sqrt[3]{xyz}\)
<=>3>=\(3\sqrt[3]{xyz}\left(dox+y+z=3\right)\)
<=>xyz<=1
<=>1/xyz>=1
<=>\(\sqrt[9]{\frac{1}{\left(xyz\right)^2}}>=1\)
do do T>=\(3\sqrt{3}\)
dau = xay ra <=>x=y=z=1
bài 1 câu b dẽ nhất
x^2 =y^4 +8
x^2 -y^4 =8
x^2 -(y^2)^2 =8
hiệu hai số cp =8
=> x =+-3 và y =+-1