Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$

$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$

$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$

$\Leftrightarrow 13x+15=2$

$\Leftrightarrow 13x=2-15=-13$

$\Leftrightarrow x=-13:13=-1$

Bài 2:

$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:

$(y+4)y=5$

$\Leftrightarrow y^2+4y-5=0$

$\Leftrightarrow (y-1)(y+5)=0$

$\Leftrightarrow y=1$ hoặc $y=-5$

Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$

Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$

a) Ta có: f(x)=-3

<=>x⁵-2x²+x⁴-x⁵+3x²-x⁴-3+2x=-3

<=>(x⁵-x⁵)+(-2x²+3x²)+(x⁴-x⁴)+2x-3=-3

<=>x²+2x-3=-3

<=>x²+2x=0

<=>x(x+2)=0

<=>x=0 hoặc x+2=0

<=>x=0 hoặc x=-2

Vậy..........

b)đa thức f(x) có nghiệm

<=>f(x)=0

<=>x²+2x-3=0

<=>x²+3x-x-3=0

<=>x(x+3)-(x+3)=0

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

<=>x-1=0 hoặc x+3=0

<=>x=1 hoặc x=-3

Vậy nghiệm của đa thức f(x) là x=-3;x=1

a, f(x)=0

b, x=1 và x=-3

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

b: \(\Leftrightarrow4x+13=5\)

hay x=-2

a) 2x(3x+1) – (2x+3)(3x-2) = 12

\(\Leftrightarrow6x^2+2x-\left(6x^2-4x+9x-6\right)=12\)

\(\Leftrightarrow6x^2+2x-6x^2+4x-9x+6=12\)

\(\Leftrightarrow-3x+6=12\) 

\(\Leftrightarrow-3x=6\)  

\(\Leftrightarrow x=-2\)  

vậy x = -2

 b)  (x+2)² – (x-3)(x+3) = 5

\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-9\right)=5\)

\(\Leftrightarrow x^2+4x+4-x^2+9-5=0\)  

\(\Leftrightarrow4x+8=0\)

\(\Leftrightarrow4x=-8\) 

\(\Leftrightarrow x=-2\)

Vậy  x = -2