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Đặt biểu thức đã cho là A.
Ta có: 2A = (3 - 1) * (3 + 1) * (3^2 + 1) * .... * (3^64 + 1)
= (3^2 - 1) * (3^2 + 1) * ... * (3^64 + 1) (hằng đẳng thức a^2 - b^ 2 = (a+b)(a-b))
Rút gọn triệt tiêu ta được 2A=3^64 - 1
=> A = (3^64 - 1)/2
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(x+1\right)^3-\left(x-4\right)\left(x+4\right)-x^3\)
\(=\left(x^3+3x^2+3x+1\right)-\left(x^2-16\right)-x^3\)
\(=x^3+3x^2+3x+1-x^2+16-x^3\)
\(=2x^2+3x+17\)
2) \(\left(x+2\right)^3-x\left(x+3\right)\left(x-3\right)-12x^2-8\)
\(=\left(x^3+6x^2+12x+8\right)-x\left(x^2-9\right)-12x^2-8\)
\(=x^3+6x^2+12x+8-x^3+9x-12x^2-8\)
\(=-6x^2+21x\)
`@` `\text {Ans}`
`\downarrow`
`1.`
\((x + 1) ^ 3 - (x - 4)(x + 4) - x ^ 3\)
`= x^3 + 3x^2 + 3x + 1 - [ x(x+4) - 4(x+4)] - x^3`
`= x^3 + 3x^2 + 3x + 1 - (x^2 + 4x - 4x - 16) - x^3`
`= x^3 + 3x^2 + 3x + 1 - (x^2 - 16) - x^3`
`= x^3 + 3x^2 + 3x + 1 - x^2 + 16 - x^3`
`= (x^3 - x^3) + (3x^2 - x^2) + 3x + (1+16)`
`= 2x^2 + 3x + 17`
`2.`
\((x + 2) ^ 3 - x(x + 3)(x - 3) - 12x ^ 2 - 8\)
`= x^3 + 6x^2 + 12x + 8 - [ (x^2 + 3x)(x-3)] - 12x^2 - 8`
`= x^3 + 6x^2 + 12x + 8 - (x^3 - 9x) - 12x^2 - 8`
`= x^3 + 6x^2 + 12x +8 - x^3 + 9x - 12x^2 - 8`
`= (x^3 - x^3) + (6x^2 - 12x^2) + (12x + 9x) + (8-8)`
`= -6x^2 + 21x `
![](https://rs.olm.vn/images/avt/0.png?1311)
à bài này dễ mà
đầu tiên nhá:không biết,tiếp theo:ko biết.Thế thôi còn lại bạn tự giải
bạn sử dụng hằng đẳng thức nhé .Mình bít nhg lười viết nắm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x^3+64}{2x^3-8x^2+32x}\\ =\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\\ =\dfrac{x+4}{2x}\)
\(\dfrac{x^3+64}{2x^3-8x^3+32x}\)
\(=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\)
\(=\dfrac{x+4}{2x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)
b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a kham khảo nha , e nhờ a e lm chứ ko phải e lm nha !
\(\left(x-2\right)\left(\frac{3}{x}+2-\frac{5}{2x}-4+\frac{8}{x^2}-4\right)\)
\(\left(x-2\right)\left[\left(\frac{3}{x}-\frac{5}{2x}\right)-6+\frac{8}{x^2}\right]\)
\(\left(x-2\right)\left(\frac{1}{2x}-6+\frac{8}{x^2}\right)\)
\(\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)\)
\(=\left(x-2\right)\left[\frac{3}{x+2}-\frac{5}{2\left(x-2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=\left(x-2\right)\left[\frac{3.2\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{8.2}{2\left(x-2\right)\left(x+2\right)}\right]\)
\(=\left(x-2\right)\left[\frac{6\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\)
\(=\left(x-2\right)\left[\frac{6\left(x-2\right)-5\left(x+2\right)+16}{2\left(x-2\right)\left(x+2\right)}\right]\)
\(=\frac{\left(x-2\right)\left(x-6\right)}{2\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-6}{2\left(x+2\right)}\)