đề bỏ số 2 nha bạn

Áp dụng BĐT Cauchy -  Schwarz, ta có :

\(\frac{a}{bc}+\frac{b}{ac}\ge2\sqrt{\frac{a}{bc}.\frac{b}{ac}}=\frac{2}{c}\)  

Tương tự , \(\frac{b}{ac}+\frac{c}{ab}\ge\frac{2}{a}\); \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\)

Cộng từng vế BĐT, ta được : 

\(2.\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\ge2.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c

Thank bạn

Biết câu nào làm câu đấy thoy nha :))

3. \(x^4y^4+4\)

\(=\left(x^2y^2\right)^2+2\cdot x^2y^2\cdot2+2^2-2\cdot x^2y^2\cdot2\)

\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)

4. \(x^4+4y^4\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot2y^2+\left(2y^2\right)^2-2\cdot x^2\cdot2y^2\)

\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)

2. \(x^4+x^2+1\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot1+1^2-2x^2\)

\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)

\(=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)\)

2. \(\left(a+b\right)^3-\left(a-b\right)^3\)

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

\(=2b\left[\left(a^2+2ab+b^2\right)+\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)

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

\(=2b\left(3a^2+b^2\right)\)

Bn cx thek nhá????

23+1+2019 =2043

kb nhoa!!

Đề sai cậu ơi !