Bài 2)

a) 2x(x - 2) + 3x - 6 = 0

2x² - 4x + 3x - 6 = 0

2x² - x - 6 = 0

2x² - 4x + 3x - 6 = 0

(2x² - 4x) + (3x - 6) = 0

2x(x - 2) + 3(x - 2) = 0

(2x + 3)(x - 2) = 0

\(\Rightarrow\left\{{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=-3\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{2}\\x=2\end{matrix}\right.\)

b) 3x(x - 5) - 5x + 25 = 0

3x(x - 5) - (5x - 25) = 0

3x(x - 5) - 5(x - 5) = 0

(3x - 5)(x - 5) = 0

\(\Rightarrow\left\{{}\begin{matrix}3x-5=0\\x-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=5\\x=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\x=5\end{matrix}\right.\)

c) x(x + 7) -x - 7 = 0

x(x + 7) - (x + 7) = 0

(x - 1)(x + 7) = 0

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

d) (5x - 20) + 3x² - 12x = 0

5(x - 4) + 3x(x - 4) = 0

(5 + 3x)(x - 4) = 0

\(\Rightarrow\left\{{}\begin{matrix}5+3x=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=-5\\x=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=4\end{matrix}\right.\)

e) (1 - x) - 3x² + 3x = 0

(1 - x) + (3x - 3x²) = 0

(1 - x) + 3x(1 - x) = 0

(1 - x)(1 + 3x) = 0

\(\Rightarrow\left\{{}\begin{matrix}1-x=0\\1+3x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\3x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)Chúc bn hok tốt

(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

(x - 1)(x - 3)(x - 5)(x - 7) - 20

= [(x - 1)(x - 7)][(x - 3)(x - 5)] - 20

= (x² - 7x - x + 7)(x² - 5x - 3x + 15) - 20

= (x² - 8x + 7)(x² - 8x + 15) - 20

Đặt x² - 8x + 7 = y, ta có:

y(y + 8) - 20

= y² + 8y - 20

= y² - 2y + 10y - 20

= (y² - 2y) + (10y - 20)

= y(y - 2) + 10(y - 2)

= (y + 10)(y - 2)

= (x² - 8x + 7 + 10)(x² - 8x + 7 - 2)

= (x² - 8x + 17)(x² - 8x + 5)

bài 4.

(2x³ - 6x² + 5x - 1) : (x - 1)

= (2x³ - 2x² - 4x² + 4x + x - 1) : (x - 1)

= [(2x³ - 2x²) - (4x² - 4x) + (x - 1)] : (x - 1)

= [2x²(x - 1)] - 4x(x - 1) + (x - 1)] : (x - 1)

= (x - 1)(2x² - 4x + 1) : (x - 1)

= 2x² - 4x + 1

3 bài trên bn có nhìn rõ ko để mk chụp gần

nếu bn ko nhìn đc thì cứ bảo mk. mk chụp lại cho

Chúc bn hok tốt

Chúc bn hok tốt

mai mk giúp cho. hôm nay mik bận làm đề cương rồi

a) (a + b)⁴

= [(a + b)²]²

= (a² + 2ab + b²)²

= [(a² + 2ab) + b²]²

= (a² + 2ab)² + 2(a² + 2ab)b² + b⁴

= a⁴ + 4a³b + 4a²b² + 2a²b² + 4ab³ + b⁴

= a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

vậy (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

con a bn chép sai đề bài nên mk sử rồi nhé

b) (a + b)⁵

= (a + b)² . (a + b)³

= (a² + 2ab + b²)(a³ + 3a²b + 3ab² + b³)

= a⁵ + 3a⁴b + 3a³b² + a²b³ + 2a⁴b + 6a³b² + 6a²b³ + 2ab⁴ + a³b² + 3a²b³ + 3ab⁴ + b⁵

= a⁵ + (3a⁴b + 2a⁴b) + (3a³b² + 6a³b²+ a³b²) + (a²b³ + 6a²b³ + 3a²b³) + (2ab⁴ 3ab⁴) + b⁵

= a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵