Ta thấy : \(\left|x-1\right|\ge0\)

\(\left|y+2007\right|\ge0\)

\(\Rightarrow B=\left|x-1\right|=2\left|y+2007\right|-2010\ge-2010\)

\(MaxB=-2010\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2007=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2007\end{cases}}}\)

a)có ng` lm r`

b)Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:

\(C-10\ge\left|x-2+2009-x\right|=2007\)

\(\Rightarrow C\ge2017\)

Dấu = khi x=2 hoặc x=2009

Vậy MinC=2017 khi x=2 hoặc x=2009

c)Xét từng trường hợp và ta có:

MinD=-1 khi \(x\ge1\)

d)\(\left|x-1\right|+\left|x-5\right|+\left|x-7\right|\)

\(\ge\left|x-1+0+7-x\right|=6\)

\(\Rightarrow E\ge6\)

Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-5=0\\x-7\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=5\\x\le7\end{cases}}\Leftrightarrow x=5\)

Vậy MinE=6 khi x=5

\(\frac{x}{9}=\frac{y}{-6}=\frac{-5x}{-45}=\frac{-3y}{18}=\frac{-5x-3y}{-27}=\frac{-81}{-27}=3\)

\(\Rightarrow\frac{x}{9}=3\Rightarrow x=27\)

\(\Rightarrow\frac{y}{-6}=3\Rightarrow y=-18\)

no no no no

Ta có:\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}\)

Ta có:\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{xa^2}{a^3}=\frac{yb^2}{b^3}=\frac{zc^2}{c^3}=\frac{a^2x+b^2y+c^2z}{a^3+b^3+c^3}\)

Ta có\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\Rightarrow\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^3}{a^2x}=\frac{y^3}{b^2y}=\frac{z^3}{c^2z}=\frac{x^3+y^3+z^3}{a^2x+b^2y+c^2z}\)

\(A=\frac{\left(x^3+y^3+z^3\right)\left(a^3+b^3+c^3\right)\left(a+b+c\right)}{\left(x+y+z\right)\left(a^2x+b^2y+c^2z\right)^2}=\frac{x^3+y^3+z^3}{a^2x+b^2y+c^2z}\cdot\frac{a^3+b^3+c^3}{a^2x+b^2y+c^2z}\cdot\frac{a+b+c}{x+y+z}\)

\(=\frac{x^2}{a^2}\cdot\frac{a}{x}\cdot\frac{a}{x}\)=1

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