Đề thiếu rồi giá cam là bao nhiêu. Sao trong đề không nói

Ta có:

\(2a+2b+2c=by+cz+ax+cz+ax+by\)

\(\Leftrightarrow a+b+c=ax+by+cz\)

\(\Rightarrow a+b+c=ax+2a;a+b+c=by+2b;a+b+c=cz+2c\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{a}{a+b+c};\frac{1}{y+2}=\frac{b}{a+b+c};\frac{1}{z+2}=\frac{c}{a+b+c}\)

\(\Rightarrow A=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

Ta có:\(\hept{\begin{cases}2a=by+cz\\2b=ax+cz\\2c=ax+by\end{cases}}\)

\(\Leftrightarrow2a+2b+2c=by+cz+ax+cz+ax+by\)

\(\Leftrightarrow2a+2b+2c=2ax+2by+2cz\)

\(\Leftrightarrow2a+2b+2c-2ax-2by-2cz=0\)

\(\Leftrightarrow\left(2a-2ax\right)+\left(2b-2by\right)+\left(2c-2cz\right)=0\)

\(\Leftrightarrow2a\left(1-x\right)+2b\left(1-y\right)+2c\left(1-z\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}1-x=0\\1-y=0\\1-z=0\end{cases}\Leftrightarrow x=y=z=1}\)

\(\Rightarrow A=\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{1+2}+\frac{1}{1+2}+\frac{1}{1+2}=1\)

x bằng 0 và y bằng 1

Ta có:

\(x^6< y^3=x^6+3x^2+1\le\left(x^2+1\right)^3\)

\(\Rightarrow x^6+3x^2+1=\left(x^2+1\right)^3\)

\(\Leftrightarrow3x^4=0\)

\(\Leftrightarrow x=0\)

\(\Rightarrow y=1\)

Đặt

\(a^2=n^2-n+2\)

Ta có:

\(\Rightarrow\left(n-1\right)^2< a^2=n^2-n+2< \left(n+1\right)^2\)

\(\Rightarrow n^2-n+2=n^2\)

\(\Leftrightarrow n=2\)

a, \(3\left(m+1\right)x+4=2x+5\left(m+1\right)\)

\(\Leftrightarrow3\left(m+1\right)x-2x=5\left(m+1\right)-4\)

\(\Leftrightarrow\left(3m+1\right)x=5m+1\)(*)

(*) có nghiệm duy nhất khi \(3m+1\ne0\Leftrightarrow m\ne-\frac{1}{3}\)

(*) vô nghiệm khi \(\hept{\begin{cases}3m+1=0\\5m+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}m=-\frac{1}{3}\\m\ne-\frac{1}{5}\end{cases}}\)

(*) đúng với mọi x khi \(\hept{\begin{cases}3m+1=0\\5m+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}m=-\frac{1}{3}\\m=-\frac{1}{5}\end{cases}}}\)

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