\(\left\{{}\begin{matrix}\left|x\right|\ge3\\\left|y\right|\ge3\\\left|z\right|\ge3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|\dfrac{1}{x}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{y}\right|\le\dfrac{1}{3}\\\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}\end{matrix}\right.\)

\(\left|A\right|=\left|\dfrac{xy+yz+xz}{xyz}\right|=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\le\left|\dfrac{1}{x}\right|+\left|\dfrac{1}{y}\right|+\left|\dfrac{1}{z}\right|\le\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\)

\(\Rightarrow A\le\left|A\right|\le1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=3\)

2)81^10-27^13-9^21=3^40-3^39-3^42=3^39(3-1-3^3) =3^39.(-25)=3^37.9.(-25)=3^37.(-225) chia hết cho 225

a²+b²+c²=ab+ac+bc

<=>2a²+2b²+2c²=2ab+2ac+2bc

<=>a²-2ab+b²+a²-2ac+c²+b²-2bc=0

<=>(a-b)²+(a-c)²+(b-c)²=0

<=>a-b=0 và a-c=0 và b-c=0

<=>a=b=c

Tl

Bạn T i k 3 lần cho mình mình trả lời cho

#Kirito

o biet

1)

Ta có : \(5-2x< 3+x\)

\(\Leftrightarrow-2x-x< 3-5\)

\(\Leftrightarrow-3x< -2\)

\(\Leftrightarrow x>\frac{2}{3}\)

Vậy bất phương trình có tập nghiệm  \(\left\{x/x>\frac{2}{3}\right\}\)

2)

Ta có : \(a^2+b^2+2-2\left(a+b\right)\)

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

\(=\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\)

\(=\left(a-1\right)^2+\left(b-1\right)^2\)

Mà : \(\left(a-1\right)^2\ge0\forall a\)

        \(\left(b-1\right)^2\ge0\forall b\)

\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2\ge0\forall a;b\) ( luôn đúng )

\(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\left(đpcm\right)\)

Vậy  \(a^2+b^2+2\ge2\left(a+b\right)\)