C18: 

\(A=a^3+b^3+c^3+3a^2+3b^2+3c^2\)'

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

\(=a\left(a+1\right)\left(a+2\right)+b\left(b+1\right)\left(b+2\right)+c\left(c+1\right)\left(c+2\right)-2\left(a+b+c\right)\) 

Xét \(a\left(a+1\right)\left(a+2\right)\) là tích 3 số nguyên liên tiếp nên tích của chúng chia hết cho 6.

Lại có \(a+b+c⋮3\) nên \(2\left(a+b+c\right)⋮6\)

Từ đó suy ra \(A⋮6\) ( đpcm )

C17.2 sai đề VD \(a=5,b=1\)

C16.6: 

\(\left(3-x\right)^4+\left(4-x\right)^4-\left(7-2x\right)^4\)

Đặt \(\left\{{}\begin{matrix}3-x=a\\4-x=b\end{matrix}\right.\)

\(a^4+b^4-\left(a+b\right)^4=a^4+b^4-a^4-b^4-4a^3b-4ab^3-6a^2b^2\)

\(=-2ab\left(2a^2+3ab+2b^2\right)=-2\left(3-x\right)\left(4-x\right)\left[2\left(3-x\right)^2+3\left(3-x\right)\left(4-x\right)+2\left(4-x\right)^2\right]\)

\(=-2\left(3-x\right)\left(4-x\right)\left(7x^2-49x+86\right)\)

C16.5: ( Dựa vào phần 1 )

\(\left(x+2\right)^4+\left(x+8\right)^4-272=\left(x+2\right)^4+\left(x+2+6\right)^4+6^4-1568\)

\(=2\left[\left(x+2\right)^2+6^2+6\left(x+2\right)\right]^2-2\cdot28^2\)

\(=2\left[\left(x+2\right)^2+6\left(x+2\right)+8\right]\left[\left(x+2\right)^2+6\left(x+2\right)+64\right]\)

\(=2\left(x+4\right)\left(x+6\right)\left[\left(x+2\right)^2+6\left(x+2\right)+64\right]\)

C16.1:\(x^4+\left(x+y\right)^4+y^4=x^4+y^4+\left(x^2+y^2+2xy\right)^2=x^4+y^4+\left(x^2+y^2\right)^2+4xy\left(x^2+y^2\right)+4x^2y^2\)

\(=2\left(x^4+y^4\right)+6x^2y^2+4x^3y+4xy^3=2\left(x^4+y^4+2x^3y+2xy^3+3x^2y^2\right)\)

\(=2\left(x^4+2x^3y+x^2y^2+2xy^3+2x^2y^2+y^4\right)\)

\(=2\left[\left(x^2+xy\right)^2+2\left(x^2+xy\right)y^2+y^4\right]=2\left(x^2+xy+y^2\right)^2\)

C15.3: 

\(3=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{3}\Leftrightarrow3\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{\sqrt[3]{abc}}\Leftrightarrow abc\ge1\)

Khi đó: \(\dfrac{1}{1+\sqrt{\left(a+b\right)^3+abc}}\le\dfrac{1}{1+\sqrt{\left(a+b\right)^3+1}}=\dfrac{1-\sqrt{\left(a+b\right)^3+1}}{1-\left(a+b\right)^3-1}=\dfrac{\sqrt{\left(a+b\right)^3+1}-1}{\left(a+b\right)^3}\)

Xét \(\sqrt{\left(a+b\right)^3+1}=\sqrt{\left(a+b+1\right)\left[\left(a+b\right)^2-\left(a+b\right)+1\right]}\le\dfrac{a+b+1+\left(a+b\right)^2-\left(a+b\right)+1}{2}=\dfrac{\left(a+b\right)^2}{2}+1\)

\(\Rightarrow\dfrac{\sqrt{\left(a+b\right)^3+1}-1}{\left(a+b\right)^3}\le\dfrac{\left(a+b\right)^2}{2.\left(a+b\right)^3}=\dfrac{1}{2}.\dfrac{1}{a+b}\le\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tương tự với 2 phân thức còn lại ta có:

\(VT\le\dfrac{1}{8}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\le\dfrac{1}{4}\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{3}{4}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

C15.2: ( Trần Văn Khắnk - Trần Thanh Fuongzz)

Theo định lý Sin: \(\dfrac{a}{sinA}=2R\Rightarrow sinA=\dfrac{a}{2R}\Rightarrow S=\dfrac{1}{2}bc.sinA=\dfrac{abc}{4R}\Leftrightarrow abc=4SR\) (1)

Gọi G là trọng tâm của tam giác ABC, ta có: 

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\Leftrightarrow3\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)

\(\Leftrightarrow9OG^2=OA^2+OB^2+OC^2+2\overrightarrow{OA}.\overrightarrow{OB}+2\overrightarrow{OB}.\overrightarrow{OC}+2\overrightarrow{OC}.\overrightarrow{OA}\)

\(\Leftrightarrow9OG^2=3R^2+2\overrightarrow{OA}.\overrightarrow{OB}+2\overrightarrow{OB}.\overrightarrow{OC}+2\overrightarrow{OC}.\overrightarrow{OA}\)

Có \(2\overrightarrow{OA}.\overrightarrow{OB}=\overrightarrow{OA}^2+\overrightarrow{OB}^2-\left(\overrightarrow{OA}-\overrightarrow{OB}\right)^2=2R^2-c^2\)

Tương tự suy ra: \(9OG^2=9R^2-\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2=9\left(R^2-OG^2\right)\) (2)

Từ (1) và (2), ta có đpcm \(\Leftrightarrow12SR\ge4S\sqrt{9\left(R^2-OG^2\right)}\)

\(\Leftrightarrow R\ge\sqrt{R^2-OG^2}\)

\(\Leftrightarrow OG^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra khi và chỉ khi \(O\equiv G\) hay tam giác ABC đều.

C15. 5: 

Áp dụng BĐT Cauchy: 

\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right).64}}=\dfrac{3a}{4}\) 

\(\Rightarrow\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\dfrac{3a}{4}-\dfrac{b+1}{8}-\dfrac{c+1}{8}\)

Tương tự: \(\Rightarrow\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\dfrac{3b}{4}-\dfrac{c+1}{8}-\dfrac{a+1}{8}\); \(\Rightarrow\dfrac{c^3}{\left(1+b\right)\left(1+a\right)}\ge\dfrac{3c}{4}-\dfrac{b+1}{8}-\dfrac{a+1}{8}\)

Cộng theo vế: \(VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{4}\left(a+b+c\right)-\dfrac{3}{4}=\dfrac{a+b+c}{2}-\dfrac{3}{4}\ge\dfrac{3\sqrt[3]{abc}}{2}-\dfrac{3}{4}=\dfrac{3}{4}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

C2 đề có vấn đề không nhỉ :v