ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)

a) 5x² ( 3x² -7x+2)-15x(x-3)

=15x⁴-35x³+10x²-15x²+45x

=15x⁴-35x³-5x²+45x

c) (x+3)(x-3)(x-2)(x+1)

=(x²-9)(x²+x-2x-2)

=(x²-9)(x²-x-2)

=x⁴-x³-2x²-9x²+9x+18

=x⁴-x³-11x²+9x+18

d)(2x+1)²+(4x-1)²+2(2x+1)(4x+1)

=2x²+4x+1-16x²-8x+1

=2x²+4x+1-16x²-8x+1+16x²-4x+8x-2

=2x²+7

e) (2x²-3x)(5x²-2x+1)-10x²(x+3)

=10x⁴ -4x³+2x²-15x³+6x²-3 -10x²-30x

=10x⁴-19x³-2x²-30x-3

thanks bn nka

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{\left(3x^3-3x^2\right)-\left(4x^2-4x\right)+\left(x-1\right)}{\left(2x^3-2x^2\right)+\left(x^2-x\right)-\left(3x-3\right)}=\frac{\left(x-1\right).\left(3x^2-4x+1\right)}{\left(x-1\right).\left(2x^2+x-3\right)}\\ \)

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

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)

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

\(=\frac{\left(3x^3-3x^2\right)-\left(4x^2-4x\right)+\left(x-1\right)}{\left(2x^3-2x^2\right)+\left(x^2-x\right)-\left(3x-3\right)}\)

\(=\frac{3x^2\left(x-1\right)-4x\left(x-1\right)+\left(x-1\right)}{2x^2\left(x-1\right)+x\left(x-1\right)-3\left(x-1\right)}\)

\(=\frac{\left(3x^2-4x+1\right)\left(x-1\right)}{\left(2x^2+x-3\right)\left(x-1\right)}\)

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

\(=\frac{3x^2-3x-x+1}{2x^2-2x+3x-3}\)

\(=\frac{\left(3x^2-3x\right)-\left(x-1\right)}{\left(2x^2-2x\right)-\left(3x-3\right)}\)

\(=\frac{3x\left(x-1\right)-\left(x-1\right)}{2x\left(x-1\right)-3\left(x-1\right)}\)

\(=\frac{3x-1}{2x-3}\)

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

a) (7x - 8)(7x + 8) - 10(2x + 3)² + 5x(3x - 2)² - 4x(x - 5)²

= 49x² - 64 - 10(4x² + 12x + 9) + 5x(9x² - 12x + 4)  - 4x(x²  - 10x + 25)

= 49x² - 64 - 40x² - 120x - 90 + 45x³ - 60x² + 20x - 4x³ + 40x - 100x

= 41x³ - 51x² - 160x - 154

b) (x2 - 3)(x² + 3) - 5x²(x + 1)² - (x² - 3x)(x² - 2x) + 4x(x + 2)²

= x⁴ - 9 - 5x2(x² + 2x + 1) - x⁴ + 5x³ - 6x² + 4x(x² + 4x + 4)

= 5x³ - 6x² - 5x⁴ - 10x³ - 5x² + 4x³ + 16x² + 16x - 9

= -5x⁴ - x³ + 5x² + 16x - 9

Trả lời:

a , ( 7x - 8 ) ( 7x + 8 ) - 10 ( 2x + 3 )² + 5x ( 3x - 2 )² - 4x ( x - 5 )²

= 49x² - 64 - 10 ( 4x² + 12x + 9 ) + 5x ( 9x² - 12x + 4 ) - 4x ( x² - 10x + 25 )

= 49x² - 64 - 40x² + 120x - 90 + 45x³ - 60x² + 20x - 4x³ + 40x² - 100x

= 41x³ - 11x² + 40x - 154

b , ( x² - 3 ) ( x² + 3 ) - 5x² ( x + 1 )² - ( x² - 3x ) ( x² - 2x ) + 4x ( x + 2 )²

= x⁴ - 9 - 5x² ( x² + 2x + 1 ) - ( x⁴ - 2x³ - 3x³ + 6x² ) + 4x ( x² + 4x + 4 )

= x⁴ - 9 - 5x⁴ - 10x³ - 5x² - x⁴ + 2x³ + 3x³ - 6x² + 4x³ + 16x² + 16x

= - 5x⁴ - x³ + 5x² + 16x - 9

\(-4x^5\left(x^3-4x^2+7x-3\right)\)

\(=-4x^8+16x^7-28x^6+12x^5\)

b) \(3x^4\left(-2x^3+5x^2-\frac{2}{3}x+\frac{1}{3}\right)\)

\(=-6x^7+15x^6-2x^5+x^4\)

đáp án câu a và b có rút gọn được không?

\(\frac{3x^3-7x^2+5x+1}{2x^3-x^2-4x+3}\)

= \(\frac{\left(3x^2-4x+1\right)\left(x-1\right)}{\left(2x^2+x-3\right)\left(x-1\right)}\)

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

= \(\frac{\left[x\left(3x-1\right)-\left(3x-1\right)\right]\left(x-1\right)}{\left[x\left(2x+3\right)-\left(2x+3\right)\right]\left(x-1\right)}\)

= \(\frac{\left(3x-1\right)\left(x-1\right)\left(x-1\right)}{\left(2x+3\right)\left(x-1\right)\left(x-1\right)}\)

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

= \(\frac{3x-1}{2x+3}\)