\(x^8+x^4+1\)

\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)

\(x^5-x^4-1\)

\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)

\(=\left(x^5-x^4+x^3\right)-\left(x^3-x^2+x\right)-\left(x^2-x+1\right)\)

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

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

L8 đã học hằng đẳng thức chưa e nhỉ?

hình như rồi

\(M=\frac{2x-1}{x^2-5x+6}=\frac{2x-1}{\left(x-2\right)\left(x-3\right)}=\frac{5\left(x-2\right)-3\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{x-3}-\frac{3}{x-2}=\frac{5}{x-3}+\frac{3}{2-x}\)

Bạn giải được câu này chưa, mình đang bí huhu

\(=\frac{2013ac}{abc+2013ac+2013c}+\frac{abc}{abc^2+abc+2013ac}+\frac{2013c}{2013ac+2013c+2013}\)

\(=\frac{2013ac}{2013+2013ac+2013c}+\frac{2013}{2013c+2013+2013ac}+\frac{2013c}{2013ac+2013c+2013}\)

\(=\frac{2013ac+2013c+2013}{2013ac+2013c+2013}=1\left(đpcm\right)\)

quy đồng

3x +6 > 6x - 2

<=> -3x > - 8

<=> x < 8/3

\(\frac{x+2}{2}>\frac{3x-1}{3}\)

<=> \(\frac{x+2}{2}-\frac{3x-1}{3}>0\)

<=> \(\frac{3\left(x+2\right)-2\left(3x-1\right)}{6}>0\)

<=> \(\frac{3x+6-6x-2}{6}>0\)

<=> \(\frac{4-3x}{6}>0\)

<=> \(4-3x>0\)

<=> \(4>3x\)

<=> \(x< \frac{4}{3}\)

BĐT cần chứng minh tương đương với :

\(\frac{a^2+b^2}{a^2+b^2+2}+\frac{b^2+c^2}{b^2+c^2+2}+\frac{c^2+a^2}{c^2+a^2+2}\ge\frac{3}{2}\)

Áp dụng BĐT Cô-si dạng Engel,ta có :

\(VT\ge\frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)}{2\left(a^2+b^2+c^2\right)+6}\)

\(\ge\frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\)

\(\ge\frac{2\left(a^2+b^2+c^2\right)+ab+bc+ac}{a^2+b^2+c^2}\ge\frac{3}{2}\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge0\)( luôn đúng )

nguồn  : loga 

Bất đẳng thức cần chứng minh tương đương: \(\Sigma\frac{2}{a^2+b^2+2}\le\frac{3}{2}\)

\(\Leftrightarrow3-\Sigma\frac{2}{a^2+b^2+2}\ge\frac{3}{2}\Leftrightarrow\Sigma\left(1-\frac{2}{a^2+b^2+2}\right)\ge\frac{3}{2}\)

\(\Leftrightarrow\Sigma\frac{a^2+b^2}{a^2+b^2+2}\ge\frac{3}{2}\)(*)

Xét vế trái của (*), ta có: \(\Sigma\frac{a^2+b^2}{a^2+b^2+2}\ge\frac{\left(\Sigma\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\)(Theo BĐT Bunyakovsky dạng phân thức)

Đến đây, ta cần chỉ ra rằng \(\frac{\left(\Sigma\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{2\left(a^2+b^2+c^2\right)+2\left(\Sigma\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\right)}{2\left(a^2+b^2+c^2\right)+6}\ge\frac{3}{2}\)\(\Leftrightarrow\frac{a^2+b^2+c^2+\Sigma\text{​​}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}}{a^2+b^2+c^2+3}\ge\frac{3}{2}\)

\(\Leftrightarrow2\text{​​}\text{​​}\Sigma\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\left(a^2+b^2+c^2\right)+9\)\(\Leftrightarrow\text{​​}\text{​​}\Sigma\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\frac{1}{2}\left(a^2+b^2+c^2\right)+\frac{9}{2}\)(**)

Theo BĐT Cauchy-Schwarz cho 2 bộ số \(\left(a;b\right)\)và \(\left(c;b\right)\), ta có:\(\left(a^2+b^2\right)\left(c^2+b^2\right)\ge\left(ac+b^2\right)^2\) \(\Rightarrow\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge ac+b^2\)(1)

Tương tự, ta có: \(\sqrt{\left(b^2+c^2\right)\left(c^2+a^2\right)}\ge ab+c^2\)(2); \(\sqrt{\left(c^2+a^2\right)\left(a^2+b^2\right)}\ge bc+a^2\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\text{​​}\text{​​}\Sigma\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge a^2+b^2+c^2+ab+bc+ca\)

\(=\frac{1}{2}\left(a^2+b^2+c^2\right)+\frac{1}{2}\left(a^2+b^2+c^2\right)+ab+bc+ca\)

\(=\frac{1}{2}\left(a^2+b^2+c^2\right)+\frac{1}{2}\left(a+b+c\right)^2=\frac{1}{2}\left(a^2+b^2+c^2\right)+\frac{9}{2}\)(Do đó (**) đúng)

Đẳng thức xảy ra khi a = b = c = 1.

Đề là gì z bạn

\(8x-3=5x+12\)

\(\Leftrightarrow8x-5x=3+12\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=15\div3\)

\(\Leftrightarrow x=5\)

P/s : Sử dụng Quy tắc chuyển vế nhé 

Ta có : 

\(8x-3=5x+12\)

\(\Rightarrow8x-5x=12+3\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=15:3\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

~ Ủng hộ nhé