1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

a, \(x^4-5x^3+2x^2+10x+2=0\)

\(\Rightarrow x^4+x^3-6x^3-6x^2+8x^2+8x+2x+2=0\)

\(\Rightarrow x^3\left(x+1\right)-6x^2\left(x+1\right)+8x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)

Vì \(x^3-6x^2+8x+2>0\) nên \(x+1=0\Rightarrow x=-1\)

Các câu còn lại tương tự!

Chúc bạn học tốt!!!

Tới đoạn \(\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)

Vì \(x^3-6x^2+8x+2\)(bấm máy tính kiểm tra nhé) khác 0 nên x+1=0

Do đó x=-1

Vậy x=-1

Chúc bạn học tốt ạNhung Phan

a. \(3x\left(2x+1\right)=6x^2+3x\)

b. \(\left(12x^3-18x^2+6x\right):6x=2x^2-3x+1\)

c. \(\dfrac{7x+6}{5x-1}+\dfrac{8x-9}{5x-1}=\dfrac{15x-3}{5x-1}=\dfrac{3\left(5x-1\right)}{5x-1}=3\)

\(a.3x\left(2x+1\right)\\ =6x^2+3x\)

\(b.\left(12x^3-18x^2+6x\right):6x\\ =2x^2-3x+1\)

\(c.\dfrac{7x+6}{5x-1}+\dfrac{8x-9}{5x-1}=\dfrac{7x+6+8x-9}{5x-1}=\dfrac{15x-3}{5x-1}=\dfrac{3\left(5x-1\right)}{5x-1}=3\)

a. 

b. 

c. \(\dfrac{7x+6}{5x-1}+\dfrac{8x-9}{5x-1}=\dfrac{15x-3}{5x-1}=\dfrac{3\left(5x-1\right)}{5x-1}=3\)

bài 1 là phân tích đa thức thành nhân tử à ?

\(15x^2-34x+15\)

\(=15x^2-25x-9x+15\)

\(=5x\left(3x-5\right)-3\left(3x-5\right)\)

\(=\left(5x-3\right)\left(3x-5\right)\)

\(x^2-25x=0\)

\(\Rightarrow x\left(x-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)

vậy_

\(\left(4x-1\right)^2-9=0\)

\(\Rightarrow\left(4x-1\right)^2-3^2=0\)

\(\Rightarrow\left(4x-1+3\right)\left(4x-1-3\right)=0\)

\(\Rightarrow\left(4x+2\right)\left(4x-4\right)=0\)

\(\Rightarrow2\cdot\left(2x+1\right)\cdot4\cdot\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}}\)

vậy_

Bài 2 :

a) \(3x^2-18x+27\)

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

\(=3\left(x^2-2\cdot x\cdot3+3^2\right)\)

\(=3\left(x+3\right)^2\)

b) \(xy-y^2-x+y\)

\(=y\left(x-y\right)-\left(x-y\right)\)

\(=\left(x-y\right)\left(y-1\right)\)

c) \(x^2-5x-6\)

\(=x^2+x-6x-6\)

\(=x\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x-6\right)\)

Bài 1.

a. x² -25x = 0

<=> x( x - 25) =0

<=> \(\orbr{\orbr{\begin{cases}x=0\\x-25=0\end{cases}}}\)

<=>\(\orbr{\begin{cases}x=0\\x=25\end{cases}}\)

b.(4x - 1)² - 9 =0

<=>16x - 8 + 1 - 9 = 0

<=>8x - 8 =0

<=> 8x = 8

<=> x = 1

Bài 2

a. 3x² - 18x +27

= 3x² - 9x - 9x + 27

= (3x² -9) - ( 9x - 27)

=3x( x - 3) - 9( x - 3)

= (x - 3)(3x - 9)

b. xy - y² - x + y

= (xy - y²) - (x - y)

= y( x - y) - (x - y)

= (x-y) (y-1)

c. x² - 5x - 6

= x² + x - 6x - 6

= (x² +x) - ( 6x + 6 )

= x( x + 1) - 6(x +1)

=(x + 1) (x - 6)

đảm bảo đúng!! 

5x.(9-x) - 15.(9-x)= 0

(9-x).(5x-15) = 0

=> 9-x = 0 => x = 9

5x -15 = 0 => 5x = 15 => x = 3

KL:...

\(\left(9-x\right)\left(5x-15\right)=0\)

\(\Leftrightarrow5\left(9-x\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=9\\x=3\end{cases}}}\)

5x(9-x)-15(9-x)=0

(9-x).(5-15)=0

(9-x).(-10)=0

-90-(-10x)=0

-90+10x=0

10x=0+90=90

x=90:10

Vậy x= 9