a) \(\left( {6{x^3} - 7{x^2} - x + 2} \right):\left( {2x + 1} \right)\)

b) $(x^4-x^3+x^2+3x):(x^2-2x+3)$

c) \(\left( {{x^2} + {y^2} + 6x + 9} \right):\left( {x + y + 3} \right)\)

\(=\left( {{x^2} + 6x + 9 - {y^2}} \right)\left( {x + y + 3} \right)\)

\(=\left[ {\left( {{x^2} + 2x.3 + {3^2}} \right) - {y^2}} \right]:\left( {x + y + 3} \right)\)

\(=\left[ {{{\left( {x + 3} \right)}^2} - {y^2}} \right]:\left( {x + y + 3} \right)\)

\(=\left( {x + 3 - y} \right)\left( {x + 3 + y} \right):\left( {x + y + 3} \right)\)

$= x + 3 - y$

$= x - y + 3$

(6x³ - 7x² - x + 2) : (2x + 1)

= (6x³ + 3x² - 10x² - 5x + 4x + 2) : (2x + 1)

= [(6x³ + 3x²) - (10x² + 5x) + (4x + 2)] : (2x + 1)

= [3x²(2x + 1) - 5x(2x + 1) + 2(2x + 1)] : (2x + 1)

= (3x² - 5x + 2)(2x + 1) : (2x + 1)

= 3x² - 5x + 2

(x⁴ - x³ + x² + 3x) : (x² - 2x + 3)

= (x⁴ + x³ - 2x³ - 2x² + 3x² + 3x) : (x² - 2x + 3)

= [(x⁴ + x³) - (2x³ + 2x²) + (3x² + 3x)] : (x² - 2x + 3)

= [x³(x + 1) - 2x²(x + 1) + 3x(x + 1)] : (x² - 2x + 3)

= (x³ - 2x² + 3x)(x + 1) : (x² - 2x + 3)

= x(x² - 2x + 3)(x + 1): (x² - 2x + 3)

= x(x + 1)

= x² + x

(x² - y² + 6x + 9) : (x + y + 3)

= [(x² + 6x + 9) - y²] : (x + y + 3)

= [(x + 3)² - y²] : (x + y + 3)

= (x + 3 + y)(x + 3 - y) : (x + y + 3)

= (x + y + 3)(x - y + 3) : (x + y + 3)

= x - y + 3

CHÚC BN HOK TỐT

c)

=

=

=

=

= x + 3 - y

= x - y + 3

1: \(\dfrac{6x^3-7x^2-x+2}{2x+2}\)

\(=\dfrac{6x^3+6x^2-13x^2-13x+12x+12-10}{2x+2}\)

\(=\dfrac{3x^2\left(2x+2\right)-\dfrac{13}{2}x\left(2x+2\right)+6\left(2x+2\right)-10}{2x+2}\)

\(=3x^2-\dfrac{13}{2}x+6-\dfrac{5}{x+1}\)

2: \(\dfrac{x^2-y^2+6x-9}{x+y+3}\)

\(=\dfrac{x^2-\left(y-3\right)^2}{x+y+3}\)

\(=\dfrac{\left(x-y+3\right)\left(x+y-3\right)}{x+y+3}\)

\(a,=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)+2\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x+2\\ b,Sửa:\left(2x^3-21x^2+67x-60\right):\left(x-5\right)\\ =\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)

a) (x³ – 7x + 3 – x²) : (x – 3)

b) (2x⁴ – 3x² – 3x² – 2 + 6x) : (x² – 2)

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\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right).\)

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

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

tui co bay max 5 nè

Chúc bạn học tốt!

\(f\left(x\right)=6x^3-7x^2-16x+m\)

Do \(f\left(x\right)\) chia hết \(2x-5\), theo định lý Bezout:

\(f\left(\dfrac{5}{2}\right)=0\Rightarrow6.\left(\dfrac{5}{2}\right)^3-7.\left(\dfrac{5}{2}\right)^2-16.\left(\dfrac{5}{2}\right)+m=0\)

\(\Rightarrow m=-10\)

Khi đó  \(f\left(x\right)=6x^3-7x^2-16x-10\)

Số dư phép chia cho \(3x-2\):

\(f\left(\dfrac{2}{3}\right)=6.\left(\dfrac{2}{3}\right)^3-7.\left(\dfrac{2}{3}\right)^2-16.\left(\dfrac{2}{3}\right)-10=-22\)

f(x)=6x3−7x2−16x+m

Do $f\left(x\right)$ chia hết $2x-5$, theo định lý Bezout:

�(52)=0⇒6.(52)3−7.(52)2−16.(52)+�=0f(25​)=0⇒6.(25​)3−7.(25​)2−16.(25​)+m=0

$\Rightarrow m=-10$

Khi đó  �(�)=6�3−7�2−16�−10f(x)=6x3−7x2−16x−10

Số dư phép chia cho $3x-2$:

�(23)=6.(23)3−7.(23)2−16.(23)−10=−22f(32​)=6.(32​)3−7.(32​)2−16.(32​)−10=−22

\(f\left(x\right)=6x^3-7x^2-16x+m\)

Do \(f\left(x\right)⋮2x-5\) , theo định lý Bezout:

\(f\left(\dfrac{5}{2}\right)=0\Rightarrow6\left(\dfrac{5}{2}\right)^3-7\left(\dfrac{5}{2}\right)^2-16\left(\dfrac{5}{2}\right)+m=0\)

\(\Rightarrow m=-10\)

Khi đó \(f\left(x\right)=6x^3-7x^2-16x-10\)

Số dư phép chia cho \(3x-2:\)

\(f\left(\dfrac{2}{3}\right)=6\left(\dfrac{2}{3}\right)^3-7\left(\dfrac{2}{3}\right)^2-16\left(\dfrac{2}{3}\right)-10=-22\)