\(3=a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\)\(abc\le1\)

\(VT=\frac{a^3\left(a+1\right)+b^3\left(b+1\right)+c^3\left(c+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a^4+b^4+c^4+a^3+b^3+c^3}{a+b+c+ab+bc+ca+abc+1}\)

\(\ge\frac{\frac{\left(a^2+b^2+c^2\right)^2}{3}+\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}}{\frac{\left(a+b+c\right)^2}{3}+5}=\frac{\frac{\frac{\left(a+b+c\right)^4}{9}}{3}+\frac{\frac{\left(a+b+c\right)^4}{9}}{3}}{8}\)

\(=\frac{\frac{\frac{3^4}{9}}{3}}{4}=\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

đề viết gì thế bạn ?

 Mk giải hộ 

 5^255 = 5^(3*75) = (5^3)^75 = 125^75 

2^572 > 2^525 
2^525 = 2^(7*75) = (2^7)^75 = 128^75 
=> 2^572 > 128^75 

128^75 > 125^75 

=> 2^572 > 2^525 > 125^75 

Đáp số: 
2^572 > 5^255

Ta có : 5⁴ > 2⁹ ( vì 625 > 512 )

=> ( 5⁴ )⁶⁴ > ( 2⁹ )⁶⁴

=> 5²⁵⁶ > 2⁵⁷⁶ 

=> 5²⁵⁶ : 5 > 2⁵⁷⁶ : 5 > 2⁵⁷⁶ : 8

=> 5²⁵⁵ > 2⁵⁷⁶ : 2³

=> 5²⁵⁵ > 2⁵⁷³ > 2⁵⁷²

=> 5²⁵⁵ > 2⁵⁷²

Vậy 5²⁵⁵ > 2⁵⁷² 

Ta có \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\left(1\right)\)

Áp dụng bất đẳng thức buniacoxki ta có :

\(\left(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\right)\left(ab+bc+ac\right)\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)^2\)

Kết hợp với (1)

=> \(\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c

Nghe mùi holder ?

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 \(\left(đpcm\right)\)