\(3x^4+11x^3-7x^2-2x+1=\left(3x^4+12x^3-3x^2-3x\right)+\left(-x^3-4x^2+x+1\right)\)

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

b,(2x -3 )(3x - 1) =(2x+3)(x-2)

<=> 6x² - 11x + 3 = 2x² - x - 6

<=.> 4x² - 10x + 9 = 0

<=> (2x - \(\frac{5}{2}\))² +\(\frac{11}{4}\)= 0 ( vô lí )

( Vì (2x - \(\frac{5}{2}\))² \(\ge\) 0 => (2x - \(\frac{5}{2}\))² + \(\frac{11}{4}\)\(\ge\)\(\frac{11}{4}\))

Vậy pt vô nghiệm

câu C : (4-3x)(2x+3)=(5-2x)(3x-4)

<=> (4-3x)(2x+3)-(5-2x)(4-3x)=0

<=>(4-3x)(2x+3-5+2x)=0

<=>(4-3x)(4x-2)=0

<=>\(\left[\begin{matrix}3x=4\\4x=2\end{matrix}\right.\)

<=>\(\left[\begin{matrix}x=\frac{4}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

d,

(3x² +1)(4x - 3 ) = ( 3x² +1 )(x - 7)

<=> (3x² + 1 )( 4x - 3) - ( 3x² + 1 )( x - 7 ) =0

<=> (3x² +1 )\(\left[\left(4x-3\right)-\left(x-7\right)\right]\)= 0

<=> (3x² + 1 ) ( 3x + 4 ) = 0

<=> 3x + 4 = 0

<=> x = -\(\frac{4}{3}\)

Vậy pt có nghiệm là - \(\frac{4}{3}\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

Lời giải:
a)

\(3x^4+11x^3-7x^2-2x+1\)

\(=(3x^4-x^3)+12x^3-4x^2-3x^2+x-3x+1\)

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

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

b)

\(x^4-6x^3+11x^2-6x+1\)

\(=(x^4-6x^3+9x^2)+2x^2-6x+1\)

\(=(x^2-3x)^2+(2x^2-6x)+1\)

\(=(x^2-3x)^2+2(x^2-3x)+1^2\)

\(=(x^2-3x+1)^2\)

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

Mình viết xuôi theo dạng ax² + bx + c nhé ;-; cho dễ làm

a) 2x² + 7x + 3 = 2x² + x + 6x + 3 = x( 2x + 1 ) + 3( 2x + 1 ) = ( 2x + 1 )( x + 3 )

b) 3x² - 8x + 4 = 3x² - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )

c) 3x² - 7x + 2 = 3x² - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )

d) -6x² + 7x - 2 = -6x² + 3x + 4x - 2 = -3x( 2x - 1 ) + 2( 2x - 1 ) = ( 2x - 1 )( 2 - 3x )

e) -3x² + 7x - 2 = -3x² + 6x + x - 2 = -3x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 1 - 3x )

f) 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 2x - 1 )

g) 3x² - 8x + 4 = 3x² - 6x - 2x + 4 = 3x( x - 2 ) - 2( x - 2 ) = ( x - 2 )( 3x - 2 )

h) 6x² - 11x + 3 = 6x² - 2x - 9x + 3 = 2x( 3x - 1 ) - 3( 3x - 1 ) = ( 3x - 1 )( 2x - 3 )

i) 2x² + 3x - 27 = 2x² - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 )

j) 4x² - 5x + 1 = 4x² - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )