\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

áp dụng cô si ta có : \(\hept{\begin{cases}a+b+c\ge3\sqrt[3]{abc}\\ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\end{cases}}\)

\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge3\cdot3\cdot\sqrt[3]{a^3b^3c^3}\)

\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\left(đpcm\right)\)

a) 1/a + 1/b + 1/c ≥ 9/(a+b+c)
<=> (1/a + 1/b + 1/c )(a+b+c) ≥ 9
Ta có : 1/a + 1/b + 1/c ≥ 3.căn bậc 3 1/abc
a+b+c ≥ 3 căn bậc 3 abc
(1/a + 1/b + 1/c)(a+c+c) ≥ 9 căn bậc 3 abc/abc = 9
<=> 1/a + 1/b + 1/c ≥ 9(a+b+c)

Hok tốt !!!!!!!!!!!

\(D=\left(x+1\right)^2+\left(2x+2\right)^2+5\)

\(=x^2+2x+1+4x^2+8x+4+5=5x^2+10x+5+5\)

\(=5\left(x^2+2x+1\right)+5=5\left(x+1\right)^2+5\ge5\)

Dấu ''='' xảy ra khi x = -1

Vậy GTNN của D bằng 5 tại x = -1

Bài 3 : 

a, Để hàm số (*) là hàm số bậc nhất khi \(m-2\ne0\Leftrightarrow m\ne2\)

b, Để hàm số (*) là hàm số đồng biến khi \(m-2>0\Leftrightarrow m>2\)

c, hàm số (*) đi qua điểm A(-2;3) <=>

 \(-2\left(m-2\right)+2m-1=3\Leftrightarrow-2m+4+2m-1=3\)( luôn đúng )

Vậy ...

\(\sqrt{\frac{1}{7}}+\frac{1}{7}\sqrt{28}-\sqrt{\left(\sqrt{7}-3\right)^2}\)

\(=\frac{\sqrt{7}}{7}+\frac{\sqrt{7.4}}{7}-\sqrt{\left(3-\sqrt{7}\right)^2}\)

\(=\frac{3\sqrt{7}}{7}-3+\sqrt{7}=\frac{3\sqrt{7}-21+7\sqrt{7}}{7}=\frac{10\sqrt{7}-21}{7}\)

\(=\frac{\sqrt{7}}{7}+\frac{2\sqrt{7}}{7}-|\sqrt{7}-3|\)   

\(=\frac{3\sqrt{7}}{7}-\left(3-\sqrt{7}\right)\)   

\(=\frac{3\sqrt{7}}{7}-3+\sqrt{7}\)

\(=\frac{3\sqrt{7}}{7}-\frac{21}{7}+\frac{7\sqrt{7}}{7}\)

\(=\frac{10\sqrt{7}-21}{7}\)

Trả lời:

\(a,\sqrt{\left(11-6\sqrt{2}\right)^2}+\sqrt{\left(11+6\sqrt{2}\right)^2}\)

\(=\left|11-6\sqrt{2}\right|+\left|11+6\sqrt{2}\right|\)

\(=11-6\sqrt{2}+11+6\sqrt{2}\)

\(=22\)

b, \(\sqrt{\left(10-4\sqrt{6}\right)^2}-\sqrt{\left(10+4\sqrt{6}\right)^2}\)

\(=\left|10-4\sqrt{6}\right|-\left|10+4\sqrt{6}\right|\)

\(=10-4\sqrt{6}-\left(10+4\sqrt{6}\right)\)

\(=10-4\sqrt{6}-10-4\sqrt{6}\)

\(=-8\sqrt{6}\)

c, \(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)

\(=\left|4-\sqrt{5}\right|+\left|1-\sqrt{5}\right|\)

\(=4-\sqrt{5}+\sqrt{5}-1\)

\(=3\)

d, \(\sqrt{\left(7+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\left|7+\sqrt{2}\right|-\left|1-\sqrt{2}\right|\)

\(=7+\sqrt{2}-\left(\sqrt{2}-1\right)\)

\(=7+\sqrt{2}-\sqrt{2}+1\)

\(=8\)

Trả lời:

Bài 2:

a, \(5\sqrt{25a^2}-25a\) với \(a\le0\)

\(=5\sqrt{\left(5a\right)^2}-25a\)

\(=5.\left|5a\right|-25a\)

\(=5.\left(-5a\right)-25a\) (vì \(a\le0\))

\(=-25a-25a=-50a\)

b, \(\sqrt{49a^2}+3a\) với \(a\ge0\) 

\(=\sqrt{\left(7a\right)^2}+3a\)

\(=\left|7a\right|+3a\)

\(=7a+3a\) (vì \(a\ge0\))

\(=10a\) 

c, \(\sqrt{16a^4}+6a^2\)

\(=\sqrt{\left(4a^2\right)^2}+6a^2\)

\(=\left|4a^2\right|+6a^2\)

\(=4a^2+6a^2=10a^2\)

d, \(3\sqrt{9a^6}-6a^3\) với \(a\le0\)

\(=3\sqrt{\left(3a^3\right)^2}-6a^3\)

\(=3.\left|3a^3\right|-6a^3\)

\(=3.\left(-3a^3\right)-6a^3\) (vì \(a\le0\))

\(=-9a^3-6a^3=-15a^3\)