Bình phương của một tổng = Bình phương của số thứ nhất + hai lần tích của số thứ nhất với số thứ hai + bình phương của số thứ hai.

\(B=1+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+\dfrac{1}{4}.\left(1+2+3+4\right)+...+\dfrac{1}{100}.\left(1+2+3+...+100\right)\)

\(B=1+\dfrac{1}{2}.2.3:2+\dfrac{1}{3}.3.4:2+\dfrac{1}{4}.4.5:2+...+\dfrac{1}{100}.100.101:2\)

\(B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{101}{2}\)

\(B=\dfrac{2+3+4+...+101}{2}\)

Tự tính :v

\(P=5x+3y+\dfrac{12}{x}+\dfrac{16}{y}\)

\(P=3x+\dfrac{12}{x+y}+\dfrac{16}{y}+2.\left(x+y\right)\)

Áp dụng BĐT Cauchy ta có:

\(3x+\dfrac{12}{x}\ge2\sqrt{\left(3.12\right)}=12\)

\(y+\dfrac{16}{y}\ge8\)

Lại có: \(2\left(x+y\right)\ge2.6=12\)

\(\Rightarrow P\ge12+8+12=32\)

Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\\x+y=6\end{matrix}\right.\)

\(\Rightarrow x=2;y=4\)

Vậy \(P_{Min}=32\Leftrightarrow\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(A=\left(x^2-2xy+y^2-12x+12y+36\right)+\left(5y^2-10y+5\right)+4\)

\(A=\left[\left(x-y\right)^2-12.\left(x-y\right)+6^2\right]+5\left(y^2-2y+1\right)+4\)

\(A=\left(x-y-6\right)^2+5.\left(y-1\right)^2+4\)

Vì \(\left(x-y-6\right)^2\ge0\forall x,y\)

\(5.\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow A_{Min}=4\Leftrightarrow y=1,x=7\)

Ta có:

\(f\left(x\right)=x^{99}+x^{88}+x^{77}+...+x^{11}+1\)

\(f\left(x\right)=\left(x^{99}+x^{88}+x^{77}+...+x^{11}\right)+1\)

\(f\left(x\right)=\left[\left(x^9\right)^{11}+\left(x^8\right)^{11}+\left(x^7\right)^{11}+...+x^{11}\right]+1\)

Ta thấy:

\(\left(x^9\right)^{11}\) chia hết cho \(x^9\)

\(\left(x^8\right)^{11}\) chia hết cho \(x^8\)

\(..........\)

\(x^{11}\) chia hết cho \(x\)

\(1\) chia hết cho \(1\)

\(\Rightarrow f\left(x\right)\) chia hết cho \(g\left(x\right)\) ( Đpcm )

CHTT

Đặt \(A=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^3}{5.7}+...+\dfrac{1006^2}{2011.2013}\)

\(\Rightarrow4A=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+\dfrac{4.3^3}{5.7}+...+\dfrac{4.1006^2}{2011.2013}\)

\(\Rightarrow4A=1006+\dfrac{1}{2}.\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right]\)

\(\Rightarrow A=\dfrac{1006+\dfrac{1}{2}.\left(1-\dfrac{1}{2013}\right)}{4}\)

\(\Rightarrow A=251,6249\)

\(a^3+a^2c-abc+b^2c+b^3=0\)

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

\(=a^2.\left(a+b+c\right)+b^2.\left(a+b+c\right)-ab^2-abc-a^2b\)

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

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

\(=0\) ( Đpcm )