cho x,y,z thoa man \(\hept{\begin{cases}xyz=2010^3\\xy+yz+zx< 2010\left(x+y+z\right)\end{cases}}\)
chung minh rang:trong 3 so co dung 1 so >2010
(goi y:dung phan chung de chung minh)
1.Giải hệ pt
1)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\xy+yz+zx=3\\\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=x\end{cases}}\)
2)\(\hept{\begin{cases}xy+yz+zx=3\\\left(x+y\right)\left(y+z\right)=\sqrt{3}z\left(1+y^2\right)\\\left(y+z\right)\left(z+x\right)=\sqrt{3}x\left(1+z^2\right)\end{cases}}\)
3)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
Chung minh rang khong co ba so x,y,z thoa man \(\hept{\begin{cases}x< y-z\\y< z-x\\z< x-y\end{cases}}\)
Bạn chỉ cần giả sử 3 số đó có tồn tại là được.
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.
cho 3 số x,y,z thỏa mãn \(\hept{\begin{cases}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{cases}}\)
tính \(P=\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)\)
\(\hept{\begin{cases}x+y+z=2010\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2010}\end{cases}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}}\)
\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+y+z\right)+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{z\left(x+z\right)+y\left(z+x\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left[\frac{\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(z+y\right)=0\)
<=> x+y = 0 hoặc x+z=0 hoặc z+y=0
<=> x = -y hoặc x = -z hoặc z = -y
\(\Rightarrow P=\left(x^{2007}+y^{2007}\right)\left(y^{2009}+z^{2009}\right)\left(z^{2009}+x^{2009}\right)=0\)
Cho x,y,z thoa mãn \(\hept{\begin{cases}x^2+xy+y^2=3\\y^2+yz+z^2=16\end{cases}}\)
Chứng minh rằng \(y^2+yz+zx\le8\)
giải hệ phương trình
a,\(\hept{\begin{cases}xy=x+3y\\yz=2\left(2y+z\right)\\zx=3\left(3z+2x\right)\end{cases}}\)
b,\(\hept{\begin{cases}x-y=3\\x^3-y^3=9\end{cases}}\)
c,\(\hept{\begin{cases}x-y=\left(\sqrt{y}-\sqrt{x}\right)\left(xy+1\right)\\x^3+y^3=54\end{cases}}\)
Em học lớp 4 thôi nên ko hiểu gì đâu ạ
\(\hept{\begin{cases}x-y=3\\\left(x-y\right).\left(x^2+xy+y^2\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=3\\x^2+xy+y^2=3\end{cases}\Leftrightarrow\hept{\begin{cases}y=x-3\\x^2+x.\left(x-3\right)+\left(x-3\right)^2=3\left(I\right)\end{cases}}}\)
Phương trình (I) tương đương: \(x^2+x^2-3x+x^2-6x+9=3\Leftrightarrow3x^2-9x+6=0\Rightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right).\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}y=-2\\y=-1\end{cases}}}\)
Vậy \(\left(x,y\right)=\left(1,-2\right),\left(2,-1\right)\)
Giải hệ pt:\(\left\{{}\begin{matrix}x^2+y^2+z^2=xy+yz+zx\\x^{2010}+y^{2010}+z^{2010}=3^{2010}\end{matrix}\right.\)
\(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
\(\Rightarrow x-y=y-z=z-x=0\)\(\Rightarrow x=y=z\)
\(\Rightarrow x^{2010}+y^{2010}+z^{2010}=3x^{2010}=3^{2010}\)
\(\Rightarrow x^{2010}=\dfrac{3^{2010}}{3}=3^{2009}\Rightarrow x=\sqrt[2010]{3^{2009}}\)
\(\Rightarrow x=y=z=\sqrt[2010]{3^{2009}}\)
Lời giải:
PT (1)
\(\Leftrightarrow x^2+y^2+z^2-(xy+yz+xz)=0\)
\(\Leftrightarrow 2(x^2+y^2+z^2)-2(xy+yz+xz)=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Thấy rằng \((x-y)^2; (y-z)^2; (z-x)^2\geq 0\forall x,y,z\in\mathbb{R}\)
\(\Rightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} (x-y)^2=0\\ (y-z)^2=0\\ (z-x)^2=0\end{matrix}\right.\Leftrightarrow x=y=z\)
Thay vào PT (2)
\(\Leftrightarrow x^{2010}+x^{2010}+x^{2010}=3^{2010}\)
\(\Leftrightarrow 3.x^{2010}=3^{2010}\Leftrightarrow x^{2010}=3^{2009}\)
\(\Leftrightarrow x=\sqrt[2010]{3^{2009}}\)
Vậy \((x,y,z)=(\sqrt[2010]{3^{2009}},\sqrt[2010]{3^{2009}},\sqrt[2010]{3^{2009}})\)
ai giup minh giai cai bai nay voi
\(\hept{\begin{cases}x^2+y^2+2x+2y=11\\xy\left(x+2\right)\left(y+2\right)=24\end{cases}}\)
voi bai \(\hept{\begin{cases}x+y+xy=1\\x+z+xz=3\\z+y+yz=7\end{cases}}\)
\(pt\left(1\right)\Leftrightarrow x\left(x+2\right)+y\left(y+2\right)=11\)
Đặt a=x(x+2); b=y(y+2) thì: \(hpt\Leftrightarrow\hept{\begin{cases}a+b=11\\ab=24\end{cases}}\)
Khi đó a,b là 2 nghiệm của pt ẩn m:
\(m^2-11m+24=0\Leftrightarrow\left(m-8\right)\left(m-3\right)=0\Rightarrow\hept{\begin{cases}m=8\\m=3\end{cases}}\)
Tới đây bn tự làm tiếp.
Giải hệ pt: \(\left\{{}\begin{matrix}x^2+y^2+z^2=xy+yz+zx\\x^{2010}+y^{2010}+z^{2010}=3^{2010}\end{matrix}\right.\)
mk nghĩ đề là \(x^{2009}+y^{2009}+z^{2009}=3^{2010}\)