Trả lời:

a, ( x² - 4x + 16 )( x + 4 ) - x ( x + 1 )( x + 2 ) + 3x² = 0

<=> x³ + 4x² - 4x² - 16x + 16x + 64 - x ( x² + 3x + 2 ) + 3x² = 0 

<=> x³ + 64 - x³ - 3x² - 2x + 3x² = 0

<=> 64 - 2x = 0

<=> 2x = 64

<=> x = 32

Vậy x = 32 là nghiệm của pt.

b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = - 50

<=> 8x - 24x² + 2 - 6x + 24x² - 60x - 4x + 10 = - 50

<=> - 62x + 12 = - 50

<=> - 62x = - 62

<=> x = 1

Vậy x = 1 là nghiệm của pt.

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

( x² - 4x + 16 )( x + 4 ) - x( x + 1 )( x + 2 ) + 3x² = 0

<=> x³ + 4³ - x( x² + 3x + 2 ) + 3x² = 0

<=> x³ + 64 - x³ - 3x² - 2x + 3x² = 0

<=> 64 - 2x = 0

<=> 2x = 64

<=> x = 32

( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = -50

<=> 2x - 24x² + 2 + 24x² - 64x + 10 = -50

<=> -62x + 12 = -50

<=> -62x = -62

<=> x = 1 

a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)

\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)

\(\Leftrightarrow9x+3=11\)

\(\Leftrightarrow9x=11-3\)

\(\Leftrightarrow9x=8\)

\(\Leftrightarrow x=\dfrac{8}{9}\)

b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)

\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)

\(\Leftrightarrow-62x+12=-50\)

\(\Leftrightarrow-62x=-50-12\)

\(\Leftrightarrow-62x=-62\)

\(\Leftrightarrow x=\dfrac{-62}{-62}\)

\(\Leftrightarrow x=1\)

a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)

\(x^2+x+3x+3-x^2+5x=11\)

\(x+8x+3=11\)

\(x+8x=8\)

\(x\left(8+1\right)=8\)

\(x=\dfrac{8}{9}\)

b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)

\(-62x+12=-50\)

\(-62x=-62\)

\(x=1\)

\(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)

\(\Leftrightarrow-62x+12=-50\)

\(\Leftrightarrow-62x=-62\)

\(\Leftrightarrow x=1\)

Có (8x + 2)(1 - 3x) + (6x - 1)(4x - 10) = -50

=>8x - 24\(^{x^2}\) + 2 - 6x + 24\(^{x^2}\) - 60x - 4x + 10 = -50

=> -62x + 12 = -50

=> -62x = -62

=> x=1 (T/m)

Vậy x=a

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)