ĐK: \(x\ne0\)

\(\frac{20x^2-15x}{5x}+\left(\frac{12-9x}{3}\right)=15\)

\(\Leftrightarrow4x-3+4-3x=15\Leftrightarrow x=14\)

\(P=x^2-4x+2x-8+9,5=x^2-2x+1-9+9,5=\)

\(=\left(x-1\right)^2+0,5>0\forall x\)

Ta có:6=2.3

         20=2².5

          15=3.5

=>BCNN(6,20,15)=2².3.5=60

=>BC(6,20,15)=B(60)={0;60;120;...}

=>x\(\in\){59:119:..}

Mà x\(\le\)30

=>x=\(\varnothing\)

Vậy x=\(\varnothing\)

Xét \(\Delta=\left(m^2+m+1\right)^2+4\left(m^2-m+1\right)>0\)

=> PT luôn có 2 nghiệm phân biệt với mọi m

Theo hệ thức Vi-et ta có \(\hept{\begin{cases}x_1+x_2=\frac{m^2+m+1}{m^2-m+1}\\x_1x_2=\frac{-1}{m^2-m+1}\end{cases}}\)

a, \(P=\frac{-1}{m^2-m+1}=\frac{-1}{\left(m-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{-1}{\frac{3}{4}}=\frac{-4}{3}\)

Dấu "=" xảy ra khi \(m=\frac{1}{2}\)

b,Tìm GTNN : lấy S trừ 2

Gợi ý:

\(A=\left|x+2\right|+\left|x-2\right|=\left|x+2\right|+\left|2-x\right|\ge\left|x+2+2-x\right|=4\)

"=" xảy ra <=> ( x+ 2 ) ( x- 2 ) \(\le0\)<=> \(-2\le x\le2\)

\(x^7+x^2+1\)

\(=x^7-x+x+x^2+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

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

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

\(=\left(x^2+x+1\right)\left[x^5-x^4+x^2-x+1\right]\)

Ta có :

\(\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)\ge\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)\left(c-\frac{1}{c}\right)\)

\(\Leftrightarrow\frac{\left(ab-1\right)\left(bc-1\right)\left(ac-1\right)}{abc}\ge\frac{\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)}{abc}\)

\(\Leftrightarrow\left(ab-1\right)\left(bc-1\right)\left(ac-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)

\(\Leftrightarrow\left(ab-bc\right)^2+\left(bc-ac\right)^2+\left(ac-ab\right)^2\ge\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

\(\Leftrightarrow\left(a-c\right)^2\left(b^2-1\right)+\left(b-c\right)^2\left(a^2-1\right)+\left(a-b\right)^2\left(c^2-1\right)\ge0\left(1\right)\)

Do a,b,c là các số thực dương không nhỏ hơn 1 nên (1) đúng .

Dấu đẳng thức xảy ra khi và khỉ khi : \(\hept{\begin{cases}\left(a-c\right)^2\left(b^2-1\right)=0\\\left(b-c\right)^2\left(a^2-1\right)=0\\\left(a-b\right)^2\left(c^2-1\right)=0\end{cases}\Rightarrow a=b=c}\)

Dấu "=" còn xảy ra ở các TH: 

a = b = 1, c bất kì .

a = c =1, b bất kì

b = c = 1,  a bất kì

( a, b, c ko nhỏ hơn 1 )