20^10>10^20

Ta có:\(10^{20}=\left(10^2\right)^{10}=100^{10}\) 

Mà \(100>20\) 

\(\Rightarrow10^{20}>20^{10}\) 

Vậy........

hc tốt

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

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

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

Bài 1 :

\(a,26+5x=3x-56\Rightarrow2x=82\Rightarrow x=41\)

\(b,x-14=3x+18\Rightarrow2x=-32\Rightarrow x=-16\)

\(c,15-\left(x+2\right)=-\left(2x+1\right)\)

\(\Rightarrow15-x-2=-2x-1\)

\(\Rightarrow x=-14\)

\(d,3\left(x-2\right)-x+2=0\)

\(\Rightarrow3\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{2;3\right\}\)

Bài 3 :

\(a,P=-a-b+c=-\left(a+b-c\right)\)là số đối của \(a+b-c\)

\(\Rightarrow\)P là số đối của Q

\(b,\left(a+b-c\right)+\left(a-b\right)-\left(a-b-c\right)\)

\(=a-b\)

Bài 4 :

\(P=a\left(b-a\right)-b\left(a-c\right)-bc.\)

\(=ab-a^2-ab+bc-bc\)\(=-a^2\)

Vì \(a^2\ge0\Rightarrow-a^2\le0\)

\(\Rightarrow P\)luôn mang giá trị âm \(\left(đpcm\right)\)

4000+300+20+1= 4321.

Ko đúng cho mk xin lỗi

Học tốt nhé bn!!!

\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}-\frac{2\left(-x-1\right)}{x+1}\right]:\frac{x-1}{x}\)

\(=\)\(\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}+\frac{2\left(x+1\right)}{x+1}\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)

\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\)\(\left(đpcm\right)\)

Áp dụng Svac + Cô-si 3 số được

\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=VP\left(đpcm\right)\)

"=" tại a = b = c