mk mới hk lớp 6

thoy mk giải lại nhá 

\(\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\left(1^2+1^2+1^2\right)>=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=1^2=1\)(bđt bunhiakopski)

dấu = xảy ra khi \(\frac{1}{x}=\frac{1}{y}=\frac{1}{z}=\frac{1}{3}\)

\(\Rightarrow3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)>=1\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}>=\frac{1}{3}\)

\(\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\left(1^2+1^2+1^2\right)>=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=1^2=1\)

dấu = xảy ra khi \(\frac{1}{x^2}=\frac{1}{y^2}=\frac{1}{z^2}=\frac{1}{3^2}=\frac{1}{9}\)

\(\Rightarrow3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)>=1\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}>=\frac{1}{3}\)

BPT\(\Leftrightarrow\left(2017x^2+2018\right)\left(2x-1\right)-\left(2017x^2+2018\right)\left(4-5x\right)\ge0\)

\(\Leftrightarrow\left(2017x^2+2018\right)\left(2x-1-4+5x\right)\ge0\)

\(\Leftrightarrow\left(2017x^2+2018\right)\left(7x-5\right)\ge0\)

DO 2017x²+2018 luôn luôn lớn hơn 0

ĐỂ B PT \(\ge\)0\(\Leftrightarrow7x-5\ge0\)

                              \(\Leftrightarrow x\ge\frac{5}{7}\)

vậy ...........

Taco:VD:2.3.4.5>24=>x<0

x=-2=>tổng=0

Vayx=-1

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)=0\)