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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: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}= \frac{x+y}{z}+1+\frac{x+z}{y}+1+ \frac{y+z}{x}+1 -3 \)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3=(x+y+z)( \frac{1}{z}+\frac{1}{x}+\frac{1}{y})-3=0-3=-3 \)
b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)
c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)
\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy \(A=0.\)
Ta có:
\(M=\frac{1}{16x}+\frac{1}{4y}+\frac{1}{z}=\frac{1}{16x}+\frac{4}{16y}+\frac{16}{16z}\)
\(\ge\frac{\left(1+2+4\right)^2}{16\left(x+y+z\right)}=\frac{49}{16}\)
Dấu bằng xảy ra khi
\(\frac{1}{16x}=\frac{2}{16y}=\frac{4}{16z}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{7}\\y=\frac{2}{7}\\z=\frac{4}{7}\end{cases}}\)
Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow\left\{{}\begin{matrix}1+\dfrac{x}{y}+\dfrac{x}{z}=0\left(1\right)\\1+\dfrac{y}{x}+\dfrac{y}{z}=0\left(2\right)\\1+\dfrac{z}{x}+\dfrac{z}{y}=0\left(3\right)\end{matrix}\right.\)
Và \(\dfrac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left(xy+yz+xz\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=0\)
\(\Rightarrow\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{y}{z}=0\)
\(\Rightarrow A+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{y}{z}=0\)
Cộng theo vế của \(\left(1\right);\left(2\right);\left(3\right)\)suy ra:
\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}=-3\)
\(\Rightarrow A-3=0\Rightarrow A=3\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
\(\Rightarrow xy=-yz-xz;yz=-xy-xz;xz=-xy-yz\)
Ta lại có: \(A=\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}=\dfrac{x^2+xy}{xz}+\dfrac{z^2+xz}{yz}+\dfrac{y^2+yz}{xy}\)
\(=\dfrac{x^2-yz-xz}{xz}+\dfrac{z^2-xy-yz}{yz}+\dfrac{y^2-xy-xz}{xy}\)
\(=\dfrac{x\left(x-z\right)}{xz}-\dfrac{yz}{xz}+\dfrac{z\left(z-y\right)}{yz}-\dfrac{xy}{yz}+\dfrac{y\left(y-x\right)}{xy}-\dfrac{xz}{xy}\)
\(=\dfrac{x-z}{z}-\dfrac{y}{x}+\dfrac{z-y}{y}-\dfrac{x}{z}+\dfrac{y-x}{x}-\dfrac{z}{y}\)
\(=\dfrac{x-z-x}{z}+\dfrac{z-y-z}{y}+\dfrac{y-x-y}{x}=\dfrac{-z}{z}+\dfrac{-y}{y}+\dfrac{-x}{x}\)
\(=-1-1-1=-3\). Vậy A=-3
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
<=> \(\dfrac{yz}{xyz}+\dfrac{xz}{xyz}+\dfrac{xy}{xyz}=0\)
<=> yz + xz + xy = 0
=> (yz)3 + (xz)3 + (xy)3 = 3x2y2z2
\(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)
= \(\dfrac{y^3z^3}{x^2y^2z^2}+\dfrac{x^3z^3}{x^2y^2z^2}+\dfrac{x^3y^3}{x^2y^2z^2}\)
= \(\dfrac{3x^2y^2z^2}{x^2y^2z^2}\)
= 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 \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow A+3=0\) \(\Rightarrow A=-3\)