giả sử x,y,z thỏa mãn x.y.z=2012
cmr:\(\frac{2012x}{xy+2012x+2012}\)\(+\frac{y}{yz+2012x+2012}\)\(+\frac{z}{xz+z+1}\)\(=1\)
cho xyz=2012. CmR \(\frac{2012x}{xy+2012x+2012}\)+\(\frac{y}{yz+y+2012}\)+\(\frac{z}{zx+y+2012}\)=1
Cho x, y, z thỏa mãn : \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}\) . Chứng minh rằng \(\frac{2012z-2013y}{2011}=\frac{2013x-2011z}{2012}=\frac{2011y-2012x}{2013}\)
Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\)
\(\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)
+) Ta có : \(\frac{2012z-2013y}{2011}=\frac{2012.2013k-2013.2012k}{2011}=0\)
\(\frac{2013x-2011z}{2012}=\frac{2013.2011k-2011.2013k}{2012}=0\)
\(\frac{2011y-2012x}{2013}=\frac{2011.2012k-2012.2011k}{2013}=0\)
Do đó : \(\frac{2012z-2013y}{2011}=\frac{2013x-2011z}{2012}=\frac{2011y-2012x}{2013}\left(=0\right)\) ( đpcm )
Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)
\(\frac{2012z-2013y}{2011}=\frac{2012\cdot2013k-2013k\cdot2012}{2011}=\frac{0}{2011}=0\)(1)
\(\frac{2013x-2011z}{2012}=\frac{2013\cdot2011k-2011\cdot2013k}{2012}=\frac{0}{2012}=0\)(2)
\(\frac{2011y-2012x}{2013}=\frac{2011\cdot2012k-2012\cdot2011k}{2013}=\frac{0}{2013}=0\)(3)
Từ (1) , (2) và (3) => đpcm
Tìm \(x\in Z\)để \(A=\frac{2013x+1}{2012x-2012}\) đạt GTLN.
Cho 3 số dương x.y.z thỏa mãn x+y+z = 1
Tìm B min = \(\frac{3}{xy+xz+yz}+\frac{2}{x^2+y^2+z^2}\)
Cho 3 số dương x,y,z thỏa mãn: xyz=1 . Tính giá trị biểu thức :
\(M=\frac{x+2xy+1}{x+xy+xz+z}+\frac{y+2yz+1}{y+yz+xy+1}+\frac{z+2xz+1}{z+xz+yz+1}\)
Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)
Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)
=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)
cho x,y,z là số thực dương thỏa mãn xy+yz+xz=xyz
cmr \(\frac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\frac{yz}{x^3\left(1+y\right)\left(1+z\right)}+\frac{xz}{y^3\left(1+x\right)\left(1+z\right)}\ge\frac{1}{16}\)
Từ \(xy+yz+xz=xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a,b,c\right)\) thì có
\(\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(a+1\right)\left(c+1\right)}+\frac{a^3}{\left(b+1\right)\left(c+1\right)}\ge\frac{1}{16}\)\(\forall\hept{\begin{cases}a+b+c=1\\a,b,c>0\end{cases}}\)
Áp dụng BĐT AM-GM ta có:
\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{64}+\frac{c+1}{64}\ge\frac{3a}{16}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế
\(VT+\frac{2\left(a+b+c+3\right)}{64}\ge\frac{3\left(a+b+c\right)}{16}\Leftrightarrow VT\ge\frac{1}{16}\)
Khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=1\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
Cho 3 số x,y,z thỏa mãn x.y.z=2019. Tính giá trị biểu thức
\(P=\frac{2019x}{xy+2019x+2019}+\frac{y}{yz+y+2019}+\frac{z}{xz+z+1}\)
\(P=\frac{2019xz}{xyz+2019xz+2019z}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{2019xz}{2019+2019xz+2019z}+\frac{y}{y\left(xz+z+1\right)}+\frac{z}{xz+z+1}\)
\(\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=1\)
Cho 3 số x;y;z khác 0 thỏa mãn xy+2013x+2013 khác 0 ; yz+y +2013 khác 0 ; xz+z+1 khác 0 và xyz=2013.
Chứng minh : \(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}=1\)
\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)
\(=\frac{xz+z+1}{xz+z+1}=1\)
=>đpcm
2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1
= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1
= xz/1+xz+z + 1/z+1+xz + z/xz+z+1
= xz+1+x/1+xz+x = 1 (đpcm)
Thay xyz=2013 vào ta có:
\(\frac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xy\cdot xz}{xy\left(xz+z+1\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz+1+z}{xz+z+1}=1\) (Đpcm)
Cho các số dương x,y,z thỏa mãn điều kiện xy + yz + xz =671
Cmr \(\frac{x}{x^2-yz-2013}+\frac{y}{y^2-xz-2013}+\frac{z}{z^2-yx-2013}\ge\frac{1}{x+y+z}\)