Chứng minh bất đẳng thức 

giải

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) \(\Leftrightarrow\dfrac{b\left(a+b\right)}{ab\left(a+b\right)}+\dfrac{a\left(a+b\right)}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\) Vì \(a,b>0\Rightarrow ab>0;a+b>0\) \(\Leftrightarrow b\left(a+b\right)+a\left(a+b\right)\ge4ab\) \(\Leftrightarrow ab+b^2+a^2+ab\ge4ab\) \(\Leftrightarrow a^2+2ab+b^2\ge4ab\) \(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\) \(\Leftrightarrow a^2-2ab+b^2\ge0\) \(\Leftrightarrow\left(a-b\right)^2\ge0\) Bất đằng thức này đúng \(\forall a,b>0\). Dấu "=" xảy ra khi \(a=b\)

HT

 (x²+x+1)(x²-x+1)(x²-1)

= x^6-1

nha bạn chúc bạn học tốt nha 

\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

\(=[\left(x-1\right)\left(x^2+x+1\right)][\left(x+1\right)\left(x^2-x+1\right)]\)

\(=\left(x^3-1\right)\left(x^3-1\right)\)

\(=\left(x^3\right)^2-1\)

\(=x^6-1\)

= ( 2x + 3 - 2x - 5 )² = (-2)² = 4

(2x+3)²-2(2x+3)(2x+5)+(2x+5)²

<=> [ (2x+3)-(2x+5) ]²

<=> (-2)²

<=> 4

 #hoctot# 

lm hết đống này á

\(-2x^2+2\left|x\right|-5\)

\(=-2\left(x^2-\left|x\right|\right)-5\)

\(=-2\left(\left|x\right|^2-2\left|x\right|\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}-5\)

\(=-2\left(\left|x\right|-\frac{1}{2}\right)^2-\frac{19}{5}\)

Mà: \(\left(\left|x\right|-\frac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow-2\left(\left|x\right|-\frac{1}{2}\right)^2\le0\)

\(\Leftrightarrow-2\left(\left|x\right|-\frac{1}{2}\right)^2-\frac{19}{5}\le\frac{-19}{5}\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|x\right|-\frac{1}{2}=0\Leftrightarrow\left|x\right|=\frac{1}{2}\Leftrightarrow x=\pm\frac{1}{2}\)

x^3y^6/8 - 27/y^3 

= -27x^(3y^6)/(8y^3)

e) \(\frac{x^3y^6}{8}-\frac{27}{y^3}=\left(\frac{xy^2}{2}\right)^3-\left(\frac{3}{y}\right)^3\)

\(=\left(\frac{xy^2}{2}-\frac{3}{y}\right)\left(\frac{x^2y^4}{4}+\frac{3xy}{2}+\frac{9}{y^2}\right)\)