\(\left(10+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Rightarrow19x-\left(10x^2+15x-2x-3\right)=8\)

\(\Rightarrow19x-10x^2-15x+2x+3=8\)

\(\Rightarrow\left(19x-15x+2x\right)-10x^2+3=8\)

\(\Rightarrow6x-10x^2+3=8\)

\(\Rightarrow6x-10x^2=5\)

mk nghĩ à đề có nhầm lẫn

(10x+9)x-(5x-1)(2x+3)=8

<=>10x²+9x-(10x²+15x-2x-3)=8

<=>10x²+9x-10x²-15x+2x+3=8

<=>-4x+3=8

<=>-4x=5

<=>x=-5/4

2x+3/5x+2 = 4x+5/10x+2

<=> (2x+3)(10x+2)=(5x+2)(4x+5)

<=>2x(10x+2)+3(10x+2)= 5x(4x+5)+2(4x+5)

<=> 20x^2+4x+20x+6 = 20x^2+25x+9x+10

<=> 20x^2+4x+20x+6 - (20x^2+25x+9x+10)=0 => 20x^2+24x+6-(20x^2+34x+10)=0

                                                                  <=>   -10x-4=0

                                                                 <=>-10x=4

                                                                  <=> x= -4/10

2x+3/5x+2=4x+5/10x+2

=> (2x+3)(10x+2)=(5x+2)(4x+5)

=> 20x^2+4x+30x+6=10x^2+25x+8x+10 ( Vì cả hai vế đều có 10x^2 nên ta xóa đi )

=> 34x+6=33x+10

=> 34x-33x=-6+10

=> x=4

x⁴+5x²-9x +3

Tìm x?

`-10x(2-x)-5x(x+2)=5x(x+3)`

`<=> -20x + 10x^2 - 5x^2 - 10x = 5x^2 +15x`

`<=> 5x^2 - 30x = 5x^2 + 15x`

`<=> -30x = 15x`

`<=> -45x = 0`

`<=> x = 0`

Vậy `S = {0}`

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

\(\text{⇔}10x\left(x-2\right)+5x\left(x-2\right)=-5x\left(x-3\right)\)

\(\text{⇔}\left(x-2\right)\left(10x+5x\right)=-5x\left(x-3\right)\)

\(\text{⇔}15x\left(x-2\right)=-5x^2+15\)

\(\text{⇔}15x^2-30=-5x^2+15\)

\(\text{⇔}15x^2+5x^2=30+15\)

\(\text{⇔}20x^2=45\)

\(\text{⇔}x=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

Ta có: \(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)

\(\Leftrightarrow-20x+10x^2-5x^2-10x-5x^2-15x=0\)

\(\Leftrightarrow-45x=0\)

hay x=0

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

=> (2x + 3)(10x + 2) = (5x + 2)(4x + 5)

=> 20x² + 4x + 30x + 6 = 20x² + 25x + 8x + 10

=> 34x + 6 = 33x + 10 (bớt 2 vế đi 20x²)

=> x = 4