\(1,M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

Thay \(a+b=1\) vào ta được:

\(1\left(1-3ab\right)+3ab\left(1-2ab\right)+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)

\(=1\)

Vậy ......................

Ta có: \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)

\(=\left(a^2+b^2+c^2\right)+\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+6\)

\(\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+6\)

\(\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{3}\left(\frac{9}{a+b+c}\right)^2+6\)

\(=\frac{100}{3}\left(đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

bn có nick fb ko mk gửi cho

thôi mk gợi ý nhé

biến đổi giả thiết như sau

(3xyz-3xy)-(3xz-3x)-(3yz-3y)+(3z-3)=x+y+z-3 =(x-1)+(y-1)+(z-1)

(=) 3(x-1)(y-1)(z-1) = (x-1)+(y-1)+(z-1)

=) 9[(x-1)(y-1)(z-1)]²=[(x-1)+(y-1)+(z-1)]² >= 3[(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)]   (áp dụng BĐT a²+b²+c²>=ab+bc+ca)

phần còn lại bn triệt tiêu 3 mỗi vế là xong

năm mới chúc bn hc tốt, chăm chỉ và nghe lời cha mẹ 

LOL GAMER   (*-*)

đáng lẽ n = 0 mới được chớ

\(ĐKXĐ:x\ne-1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow x-2-5x-5=15\)

\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)

Vậy \(S=\left\{\frac{-11}{2}\right\}\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)

\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow-4x-7=15\)

\(\Leftrightarrow-4x=22\)

\(\Leftrightarrow x=22:\left(-4\right)\)

\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)

Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)