Thực hiện phép chia ta được thương là: \(2x^2+2x+1\)

Đặt \(A=2x^2+2x+1=2\left(x^2+x+\frac{1}{4}\right)+\frac{1}{2}=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall x\)

Dấu "=" xảy ra khi: \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

Chúc bạn học tốt.

Cảm ơn Pham Van Hung nhé😆🙋

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

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

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

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

\(x^7+x^5+1\)

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

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

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

\(x^8+x^2+1>0\) \(\forall x\)

=> ko thể phân tích thành nhân tử

\(b^5+b+1\)

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

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

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

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

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

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

\(=2-6xy-3+6xy\)

\(=-1\)

\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

=>  \(A+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)

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

\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(\ge\frac{1}{2}.3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.3\sqrt[3]{\frac{1}{a+b}.\frac{1}{b+c}.\frac{1}{c+a}}=\frac{9}{2}\)   (AM - GM)

=>  \(A\ge\frac{9}{2}-3=\frac{3}{2}\)  (đpcm)

Đặt \(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(A=\frac{a^2}{ba+ca}+\frac{b^2}{cb+ba}+\frac{c^2}{ac+bc}\)

Áp dụng BĐT Cauchy-schwarz ta có: 

\(A=\frac{a^2}{ba+ca}+\frac{b^2}{cb+ba}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2.\left(ab+bc+ca\right)}\)

Ta c/m BĐT phụ \(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)( tự c/m)

Áp dụng: 

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

                                                 đpcm

Tham khảo nhé~

Con tham khảo bài tương tự tại đây nhé:

Câu hỏi của ngoc Ngoc - Toán lớp 7 - Học toán với OnlineMath

\(\frac{5z-6y}{4}\)=\(\frac{6x-4y}{5}=\frac{4y-5x}{6}=\frac{20z-24y}{16}=\frac{30x-20z}{25}=\frac{24y-30x}{36}\)

=\(\frac{20z-24y+30x-20z+24y-30x}{16+25+36}=\frac{0}{77}=0\)

=>\(\frac{5z-6y}{4}=0=>5z-6y=0=>5z=6y=>\frac{y}{5}=\frac{z}{6}\left(1\right)\)

Tương tự ta có:\(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}\)

*Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{3z}{12}=\frac{2y}{10}=\frac{5z}{30}=\frac{3z-2y+5z}{12-10+30}=\frac{96}{32}=3\)

Tự giải nhé Đô Long

                                                              Kí tên: BTS V