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ta có:\(\frac{x-y}{z-y}=-10\)
<=>\(x-y=10y-10z\)
<=>\(11y=-\left(x+10z\right)\)
<=>\(11y-11z=-\left(x-z\right)\)
<=>\(x-z=-\frac{11\left(y-z\right)}{ }\)
tahy vào biểu thức thì GT bằng -11
dap an A Tam An 2A6 tieu hoc thanh xuan hoc thanh xuan bac giu tin nhan
Bài 1
\(a^2-2a+6b+b^2=-10\)
<=>\(a^2-2a+1+b^2+6b+9=0\)
<=>\((a-1)^2+(b+3)^2=0\)
Ta lại có: \((a-1)^2\ge0 \)
\((b+3)^2\ge0\)
=> \((a-1)^2+(b+3)^2\ge0\)
Mà\((a-1)^2+(b+3)^2=0\)
=>(a-1)2=0=>a=1
(b+3)2=0=>b=-3
Vậy a=1,b=-3
Bài 2
Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}= \frac{x+y}{z}+1+\frac{x+z}{y}+1+ \frac{y+z}{x}+1 -3 \)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3=(x+y+z)( \frac{1}{z}+\frac{1}{x}+\frac{1}{y})-3=0-3=-3 \)
Ta có: \(A=\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}\)
\(\Rightarrow A+3=\dfrac{x+y}{z}+1+\dfrac{x+z}{y}+1+\dfrac{y+z}{x}+1\)
\(=\dfrac{x+y+z}{z}+\dfrac{x+y+z}{y}+\dfrac{x+y+z}{x}\)
\(=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Mà \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow A+3=0\) \(\Rightarrow A=-3\)
Áp dụng bđt Svác - sơ ta có :
\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{\left(x+y+z\right)^2}{2.\dfrac{\left(x+y+z\right)^2}{3}}=\dfrac{3}{2}\) có GTNN là \(\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
vi a/x + b/y + c/z =0 suy ra ayz/xyz + bxz/xyz + cxy/xyz =0 suy ra ayz+bxz+cxy /xyz =0 suy ra ayz + bxz + cxy =0
vi x/a + y/b =z/c =0 suy ra (x/a + y/b + z/c )^2 =0 suy ra x^2/a^2 +y^2/b^2 + z^2/c^2 + 2(xy/ab + xz/ac + yz/bc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(cxy+ bxz +ayz /abc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 +2011 = 2011
Ta sẽ CM BĐT phụ sau : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Áp dụng BĐT Cauchy dang Engel , ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{3^2}{a+b+c}=\dfrac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Trong đó : \(\left\{{}\begin{matrix}a=x+y\\b=y+z\\c=z+x\end{matrix}\right.\) , ta có :
\(\left(x+y+y+z+x+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)\ge9\)
\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)\ge4,5\)
\(\Leftrightarrow\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}\ge4,5\)
\(\Leftrightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}\ge4,5\)
\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{z+y}\ge1,5\)
\(\Rightarrow P_{Min}=1,5."="\Leftrightarrow x=y=z\)
Ta có: \(\dfrac{x-y}{z-y}=-10\)
nên \(z-y=\dfrac{x-y}{-10}\)
hay \(y-z=\dfrac{x-y}{10}=\dfrac{1}{10}\left(x-y\right)\)
Ta có: \(\dfrac{x-y}{z-y}=-10\)
\(\Leftrightarrow\dfrac{x-y}{-10}=\dfrac{z-y}{1}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y}{-10}=\dfrac{z-y}{1}=\dfrac{x-y-z+y}{-10-1}=\dfrac{x-z}{-11}\)
Do đó: \(\dfrac{x-y}{-10}=\dfrac{x-z}{-11}\)
\(\Leftrightarrow x-z=\dfrac{11\left(x-y\right)}{10}=\dfrac{11}{10}\left(x-y\right)\)
\(\Leftrightarrow\dfrac{x-z}{y-z}=\dfrac{11}{10}\left(x-y\right):\dfrac{1}{10}\left(x-y\right)=\dfrac{11}{10}\cdot\dfrac{10}{1}=11\)