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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
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
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 `
a) \(\frac{x^2+2x+4}{4x^3-32}=\frac{x^2+2x+4}{4\left(x^3-8\right)}=\frac{x^2+2x+4}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{1}{4\left(x-2\right)}.\)
b) \(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}.\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
HAND!!!!
\(\frac{x^2+2x+4}{4x^3-32}=\frac{\left(x+2\right)^2}{4\left(x^3-8\right)}=\frac{\left(x+2\right)^2}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{x+2}{4\left(x^2+2x+4\right)}.\)
\(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}\)
RÚT GỌN BIỂU THỨC
a) 32(x+2)(x-2)-1/2(6-8x)2 -48
b) (x+9)(x2+27)-(x+3)3
c) (6x+1)2(6x-1)2-2(1+6x)(6x-1)
a) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=\left(x^2-1\right)\left[\left(x^2-1\right)^2-\left(x^4+x^2+1\right)\right]\)
\(=\left(x^2-1\right)\left(x^4-2x^2+1-x^4-x^2-1\right)=\left(x^2-1\right)\left(-3x^2\right)\)
\(=-3x^4+3x^2=3\left(x^2-x^4\right)=3\left(x-x^2\right)\left(x+x^2\right)=\left(3x-3x^2\right)\left(x+x^2\right).\)
b)\(\left(x^4-3x^2+9\right)\left(x^2+3-\left(3+x^2\right)\right)^3=\left(x^4-3x^2+9\right).0^3=0\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=\left(x-3\right)^3-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)\)
\(=\left(x-3\right)^3-\left[\left(x-3\right)^3+3.x.3.\left(x-3\right)\right]+6x^2+12x+6\)
\(=6x^2+12x+6-9x\left(x-3\right)=6x^2+12x+6-9x^2+27x\)
\(=39x-3x^2+6=3\left(13x-x^2+2\right).\)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
b) Điều kiện : x ≠ -1 và x ≠ 2