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\(\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+y+z+1}+\frac{zx+2z+1}{zx+z+x+1}\)
Ta có: \(\frac{xy+2x+1}{xy+x+y+1}=\frac{\left(xy+x\right)+\left(x+1\right)}{\left(xy+x\right)+\left(y+1\right)}=\frac{x\left(y+1\right)+\left(x+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{x}{x+1}+\frac{1}{y+1}\)
Tương tự ta có:
\(\frac{yz+2y+1}{yz+y+z+1}=\frac{y}{y+1}+\frac{1}{z+1}\)
\(\frac{zx+2z+1}{zx+z+x+1}=\frac{z}{z+1}+\frac{1}{x+1}\)
Từ đây ta có biểu thức ban đầu sẽ bằng
\(\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}\)
\(\left(\frac{x}{x+1}+\frac{1}{x+1}\right)+\left(\frac{y}{y+1}+\frac{1}{y+1}\right)+\left(\frac{z}{z+1}+\frac{1}{z+1}\right)=1+1+1=3\)
CHÚ Ý: ab+a+b+1=a(b+1)+(b+1)=(a+1)(b+1)
Xét: \(\frac{xy+2x+1}{xy+x+y+1}=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}=\frac{x}{x+1}+\frac{1}{y+1}\)
Tương tự với 2 biểu thức còn lại ta được:
A=\(\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}\)
=\(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}=1+1+1=3\)
#)Góp ý :
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tk cho mk nhé
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
b)\(\frac{x^2-25}{5x-x^2}\)=\(\frac{\left(x-5\right)\left(x+5\right)}{x\left(5-x\right)}\)=\(\frac{\left(x-5\right)\left(x+5\right)}{-x\left(x-5\right)}\)=\(\frac{-x-5}{x}\)
c)\(\frac{y^2-xy}{4xy-4y^2}\)=\(\frac{y\left(y-x\right)}{4y\left(x-y\right)}=\frac{-y\left(x-y\right)}{4y\left(x-y\right)}=\frac{-1}{4}\)
d)\(\frac{x^2+xz-xy-yz}{x^2+xz+xy+yz}=\frac{x\left(x+z\right)-y\left(x+z\right)}{x\left(x+z\right)+y\left(x+z\right)}=\frac{\left(x+z\right)\left(x-y\right)}{\left(x+z\right)\left(x+y\right)}=\frac{x-y}{x+y}\)
lẽ ra x,y,z>0 chứ sao lại a,b,c>0 :))
Áp dụng bđt Cô-si:\(x^2+yz\ge2\sqrt{x^2.yz}=2x\sqrt{yz}\Leftrightarrow\frac{1}{x^2+yz}\le\frac{1}{2x\sqrt{yz}}\)
tương tự: \(\frac{1}{y^2+xz}\le\frac{1}{2y\sqrt{xz}};\frac{1}{z^2+xy}\le\frac{1}{2z\sqrt{xy}}\)
=>\(\frac{1}{x^2+yz}\)\(+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\le\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2xyz}\)
Mặt khác theo bđt Cô-si thì: \(\sqrt{xy}\le\frac{x+y}{2};\sqrt{yz}\le\frac{y+z}{2};\sqrt{xz}\le\frac{x+z}{2}\)
=>\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}=\frac{2\left(x+y+z\right)}{2}=x+y+z\)
=>\(\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\le\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2xyz}\le\frac{x+y+z}{2xyz}=\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
ta có đpcm.
Áp dụng cauchy cho mỗi mẫu số vế trái , có :
\(VT\le\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{1}{2}.\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\)
\(=\frac{1}{2}.\left(\frac{\sqrt{yz}}{xyz}+\frac{\sqrt{xz}}{xyz}+\frac{\sqrt{zx}}{xyz}\right)=\frac{1}{2}.\frac{\sqrt{yz}+\sqrt{xz}+\sqrt{xz}}{xyz}\)
Biến đổi vế phải , có :
\(VP=\frac{1}{2}.\left(\frac{z}{xyz}+\frac{x}{xyz}+\frac{y}{xyz}\right)=\frac{1}{2}.\frac{x+y+z}{xyz}\)
Ta có :
\(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
<=> \(2x+2y+2z\ge2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}\) (đúng - Hệ quả của Cauchy, lên mạng sợt là ra )
=> \(\frac{1}{2}.\frac{x+y+z}{xyz}\ge\frac{1}{2}.\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\)
=> \(VP\ge VT\)
Ta có: \(P=\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}=\frac{1}{\frac{x}{\sqrt{yz}}+2}+\frac{1}{\frac{y}{\sqrt{zx}}+2}+\frac{1}{\frac{z}{\sqrt{xy}}+2}\)
Đặt \(\frac{x}{\sqrt{yz}}=c,\frac{y}{\sqrt{zx}}=t;\frac{z}{\sqrt{xy}}=k\left(c,t,k>0\right)\)thì ctk = 1
Ta cần tìm giá trị lớn nhất của \(P=\frac{1}{c+2}+\frac{1}{t+2}+\frac{1}{k+2}\)với ctk = 1
Dự đoán MaxP = 1 khi c = t = k = 1
Thật vậy: \(P=\frac{kt+2k+2t+4+ct+2c+2t+4+ck+2c+2k+4}{\left(c+2\right)\left(t+2\right)\left(k+2\right)}=\frac{\left(kt+tc+ck\right)+4\left(c+t+k\right)+12}{ctk+2\left(kt+tc+ck\right)+4\left(c+t+k\right)+8}\le\frac{\left(kt+tc+ck\right)+4\left(c+t+k\right)+12}{1+\left(kt+tc+ck\right)+3\sqrt[3]{\left(ctk\right)^2}+4\left(c+t+k\right)+8}=1\)Đẳng thức xảy ra khi x = y = z
Ta có: \(\frac{\sqrt{yz}}{x+2\sqrt{yz}}=\frac{1}{2}\left(1-\frac{x}{x+2\sqrt{yz}}\right)\le\frac{1}{2}\left(1-\frac{x}{x+y+z}\right)=\frac{1}{2}\left(\frac{y+z}{x+y+z}\right)\)(bđt cosi) (1)
CMTT: \(\frac{\sqrt{xz}}{y+2\sqrt{xz}}\le\frac{1}{2}\left(\frac{x+z}{x+y+z}\right)\)(2)
\(\frac{\sqrt{xy}}{z+2\sqrt{xy}}\le\frac{1}{2}\left(\frac{x+y}{x+y+z}\right)\)(3)
Từ (1), (2) và (3) cộng vế theo vế ta có:
\(\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{xz}}{y+2\sqrt{xz}}+\frac{\sqrt{xy}}{z+2\sqrt{xy}}\le\frac{1}{2}\left(\frac{y+z}{x+y+z}\right)+\frac{1}{2}\left(\frac{x+z}{x+y+z}\right)+\frac{1}{2}\left(\frac{x+y}{x+y+z}\right)\)
=> P \(\le\frac{1}{2}\left(\frac{y+z+x+z+x+y}{x+y+z}\right)=\frac{1}{2}\cdot\frac{2\left(x+y+z\right)}{x+y+z}=1\)
Dấu "=" xảy ra <=> x = y = z
Vậy MaxP = 1 <=> x = y = z
a) \(\frac{3m-6n}{10n-5m}\)
\(=\frac{-3\left(2n-m\right)}{5\left(2n-m\right)}=\frac{-3}{5}\)
b) \(\frac{y^3+y^2+4y+4}{y^2+2y-8}\)
\(=\frac{y^2\left(y+1\right)+4\left(y+1\right)}{y^2+2y+1-9}\)
\(=\frac{\left(y^2+4\right)\left(y+1\right)}{\left(y+1\right)^2-9}\)
\(=\frac{\left(y^2+4\right)\left(y+1\right)}{\left(y-2\right)\left(y+4\right)}\)
c) \(\frac{x^2-xy-xz+yz}{x^2+xy-xz-yz}\)
\(=\frac{x\left(x-y\right)-z\left(x-y\right)}{x\left(x+y\right)-z\left(x+y\right)}\)
\(=\frac{\left(x-z\right)\left(x-y\right)}{\left(x-z\right)\left(x+y\right)}\)
\(=\frac{x-y}{x+y}\)