1) (2x^2 + 1)(x^2 - 2x - 1)

= 2x^4 - 4x^3 - 2x^2 + x^2 - 2x - 1

= 2x^4 - 4x^3 - x^2 - 2x - 1

2) (x^2 - x^4)/(x^2 - 1 + 1)

= (x^2.(1 - x^2))/(x^2 - 1 + 1)

= (x^2.(1 + x)(1 - x))/x^2

= (1 + x)(1 - x)

3) (3x + y)^3 + x^3 - 3x^2 + 3x + 1

Thay x = 1,1; y = -0,7 vào biểu thức, ta có:

= [3.1,1 + (-0,7)]^3 + 1,1^3 - 3.1,1^2 + 3.1,1 + 1

= 19,577

a) thay x = -3 vào biểu thức, ta có: 

\(A=\frac{\left(-3\right)^2+2.\left(-3\right)}{\left(-3\right)+1}=-\frac{3}{2}\)

b) M = A.B

\(M=\left(-\frac{3}{2}\right)\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)\)

\(M=-\frac{3\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)}{2}\)

\(M=-\frac{3.\frac{8}{x+2}}{2}\)

\(M=-\frac{\frac{24}{x+2}}{2}\)

\(M=-\frac{24}{2\left(x+2\right)}\)

\(M=-\frac{12}{x+2}\)

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

\(\Rightarrow2A=8.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

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

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

.....

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

\(=3^{128}-1\)

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

a, ABD đồng dạng ACE (g.g) (có chung góc A và cùng có 1 góc vuông)

b, từ câu a => AD/AB = AE/AC

2 tam giác có chung góc A => đồng dạng  (c.g.c)

\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}\right)=\left(x-23\right)\left(\frac{1}{26}+\frac{1}{27}\right)\text{ nhận thấy:}\frac{1}{24}+\frac{1}{25}>\frac{1}{26}+\frac{1}{27}\)

\(\Rightarrow x-23=0\Leftrightarrow x=23\)

\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\Rightarrow\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)=\left(\frac{x+3}{2002}+1\right)+\left(\frac{x+4}{2001}+1\right)\)

\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\text{dạng giống câu a rồi nha}\)

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\Leftrightarrow300-x=0\)

Vậy: x=300