Ta có \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^{64}-1\right)\left(3^{64}+1\right)=\left(3^{128}-1\right)\)

\(\Rightarrow A=\frac{3^{128}-1}{2}\)

Ta có 

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

\(=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36\)

\(=10x^2+40x+50\)

\(=x^2+10x+25+9x^2+30x+25\)

\(=\left(x+5\right)^2+\left(3x+5\right)^2\) (đpcm)

\(ĐKXĐ:x\ne0;x\ne-3;x\ne-6;x\ne-9\)

\(\frac{1}{x^2+3x}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+9}=\frac{9}{10}\)

\(\Leftrightarrow\frac{9}{x\left(x+9\right)}=\frac{9}{10}\)

\(\Leftrightarrow x^2+9x-10=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-10\end{cases}\left(tm\right)}\)

O.O chắc ko 

\(\frac{1}{x^2+3x}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)\(ĐKXĐ:x\ne-3;-6\)

\(\frac{1}{x\left(x+3\right)}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)

\(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{x^2+15x+54}=\frac{3}{10}\)

\(10\left(x+6\right)\left(x+9\right)+10x\left(x+9\right)+10x\left(x+3\right)=3x\left(x+3\right)\left(x+6\right)\left(x+9\right)\)

\(30x^2+270x+540=3x^4+54x^3+297x^2+486x\)

\(30x^2+270x+540-3x^4-54x^3-297x^2-486x=0\)

\(-3\left(89x^2+72x-180+x^4+18x^3\right)=0\)

\(-3\left(x^2+16x+60\right)\left(x-1\right)=0\)

\(-3\left(x+6\right)\left(x+10\right)\left(x+3\right)\left(x-1\right)=0\)

\(\left(x+6\right)\left(x+10\right)\left(x+3\right)\left(x-1\right)=0\)

\(x=-10,1\)

a, CM: AD//AB=AE//AC

Xét tam giác ABC có:

AD//AB vì đề bài cho cạnh BC lấy D ( lấy sao cho AD=AB)

AE//AC vì đề bài cho cạnh AC lấy E  ( lấy sao cho AE=AC)

VÌ ĐỀU CHUNG MỘT TAM GIÁC NÊN 3 CẠNH = NHAU 

\(\Rightarrow\) AD/AB=AE/AC.

b, AB = 2cm vì AD= 2cm( AD//AB \(\Rightarrow=\)nhau và = 2 cm)

đêm hôm khuya khoắt đăng lên lm j :v 

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

\(=\left(a+b+c-a\right)\left(3x^2+b^2+c^2+3ab+2bc+3ac\right)-\left(b^3+c^3\right)\)

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

\(=\left(3a^2+3ab+3ac+3bc\right)\left(b+c\right)\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

\(\text{ (a+b+c)^3−a^3−b^3−c^3 =a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^2−a^3−b^3−c^3 =3(b+c)(a^2+ab+ac)+b^3+3b^2c+3bc^2+c^3−b^3−c^3 =3(b+c)(a^2+ab+ac+bc) =3(b+c)[a(a+b)+c(a+b)] =3(b+c)(a+b)(a+c)}\)

2X hay 2x ?

\(\frac{2x-1}{x^2+1}=\frac{2}{x^2-x+1}-\frac{1}{x+1}\)đề như này hả , hay thế nào == 

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

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

\(2x^4-x^3-2x-1-3x^3-3x-1+x^4=0\)

\(3x^4-4x^3-x-2=0\)

\(x=-0.618034;1.618034\)( đù -.- ra lắm :v ) 

Ta có : \(x.\left(x+1\right)-2x=0\)

\(\Leftrightarrow2x+x-2x=0\)

\(\Leftrightarrow x=0\)

Vậy x=0 

hok tốt!!

x(x-1)-2x=0

x^2-x-2x=0

x^2-3x=0

x(x-3)=0

\(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)