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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=x^3-27-x^3-27x+x+27=-26x\)
b: \(=x^2-14x-10x^2+20x-10=-9x^2+6x-10\)
c: \(\Leftrightarrow2x^2-4x-4x^2-6x+2x+3=0\)
=>3=0(vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`
`=x^3-3^3-(9x^3+27x^2-9x-27)`
`=x^3-3^3-9x^3-27x^2+9x+27`
`=-8x^3-27x^2+9x`
b, `(x-2)(x^2+2x+4)-x(x-3)(x+3)`
`=x^3-2^3-x(x^2-9)`
`=x^3-8-x^3+9x`
`=9x-8`
a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(9x+27\right)\)
\(=x^3-27-\left(9x^3+27x^2-9x-27\right)\)
\(=x^3-27-9x^3-27x^2+9x+27\)
\(=-8x^3-27x^2+9x\)
b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=x^3-8-x\left(x^2-9\right)\)
\(=x^3-8-x^3+9x\)
\(=9x-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn đăng từng bài 1 và tách bài ra cho dễ nhìn hơn nhé!
3A:
a: =15x^4-5x^2-24x^4+18x^2-6x-6x^4+2x^3
=-15x^4+2x^3+13x^2-6x
b: =1/2(x^3-2/5x^2+2x)-3/4x^3-1/4x^2-x^2-x
=1/2x^3-1/5x^2+x-3/4x^3-5/4x^2-x
=-1/4x^3-29/20x^2
c: =3/2x^2(x^2-2x)-2x(x^3+x^2+1)+2(x-1)
=3/2x^4-3x^3-2x^4-2x^3-2x+2x-2
=-1/2x^4-5x^3-2
d: =x^4-2x^3+5x^3-10x^2+5/2x-x^4+x^3-x^2
=4x^3-11x^2+5/2x
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)
\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)
\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)
\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)
a) (x-3)(x2+3x+9)-(x2-1)(x+27)
=(x3-27)-(x3+27x2-x-27)
=x3-27-x3-27x2+x+27
=-27x2+x
=x(-27x+1)
b) (3-x)3-(x+3)(x2-3x+9)
=27-27x+9x2-x3-x3-27
=-2x3+9x2-27x
=x(-2x+9x-27)
c) (x-2)(x2+2x+4)-x(x-3)(x+3)
=x3-8-x(x2-9)
=x3-8-x3+9x
=9x-8
#H
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (2x - 1)(x + 2) - 3x² + (x - 1)²
= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1
= (2x² - 3x² + x²) + (4x - x - 2x) + (-2 + 1)
= x - 1
B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)
= x³ - 8 - x³ - x² - 9 + x²
= (x³ - x³) + (-x² + x²) + (-8 - 9)
= -17
A = (2x - 1)(x + 2) - 3x² + (x - 1)²
= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1
= (2x² - 3x² + x²) + (4x - x - 2x) + (-2 + 1)
= x - 1
B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)
= x³ - 8 - x³ - x² - 9 + x²
= (x³ - x³) + (-x² + x²) + (-8 - 9)
= -17
kết quả đây
chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
a) Ta có: \(\left(x-2\right)^3-\left(3+x^2\right)\left(3-x\right)\)
\(=x^3-6x^2+12x-8+\left(x-3\right)\left(x^2+3\right)\)
\(=x^3-6x^2+12x-8+x^3+3x-3x^2-9\)
\(=2x^3-9x^2+15x-17\)
b) Ta có: \(x\left(x-14\right)-10\left(x-1\right)^2\)
\(=x^2-14x-10\left(x^2-2x+1\right)\)
\(=x^2-14x-10x^2+20x-10\)
\(=-9x^2+6x-10\)
c) Ta có: \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)
\(=2x^2+4x-\left(x^2-4\right)\)
\(=2x^2+4x-x^2+4\)
\(=x^2+4x+4\)
d) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3-27\right)\)
\(=x^3-27-x^3+27\)
=0