a, \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)

\(=2x.\left(4x+2\right)=4x\left(2x+1\right)\)

b, \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)

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

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Tên cũng được, chưa đến nỗi !

a) Số vừa chia hết cho 2 vừa chia hết cho 5 là: 480;2000;9010/

b) Số chia hết cho 2 nhưng không chia hết cho 5 là: 296;324.

c) Số chia hết cho 5 nhưng không chia hết cho 2 là: 345;3995

Tick nhoa?

\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)

\(\Leftrightarrow x^3-12x^2+48x-64-x^3+125=x^3+8-x^3-12x^2-48x+64\)

\(\Leftrightarrow-12x^2+48x+61=-12x^2-48x+72\)

\(\Leftrightarrow48x+61=-48x-72\)

\(\Leftrightarrow x=\dfrac{-133}{96}\)

Đặt \(A=1.2+2.3+...+19.20\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+19.20.\left(21-18\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+19.20.21-18.19.20\)

\(\Rightarrow3A=19.20.21\)

\(\Rightarrow A=19.20.7=2660\)

Vậy A = 2660

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(B=\left|x-102\right|+\left|2-x\right|=\left|102-x\right|+\left|x-2\right|\ge\left|x-102+2-x\right|=100\)

Dấu " = " khi \(\left\{{}\begin{matrix}102-x\ge0\\x-2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le102\\x\ge2\end{matrix}\right.\)

Vậy \(MIN_B=100\) khi \(2\le x\le102\)

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

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

chứng minh j?

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

\(=-\left(x^2-4x+4-8\right)=-\left[\left(x-2\right)^2-8\right]\)

\(=-\left(x-2\right)^2+\le8\)

Dấu " = " khi \(-\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy \(MAX_B=8\) khi x = 2

b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{33}\)

\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)

\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)

\(=\dfrac{92}{5005}\)

\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)

Vậy...