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: $x^3+y^3=(-1)^3+(-5)^3=-126$

\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)

\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)

\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)

\(\Rightarrow-14x-60=0\)

\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)

\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)

\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)

\(\Rightarrow-x^2+12x-32=7x-x^2-10\)

\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)

\(\Rightarrow5x-22=0\)

\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)

a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1

15x + 25 - 8x + 12 = 5x + 16x + 96 + 1

15x - 8x - 5x - 16x = 96 + 1 - 25 - 12

-14x = 60

x = \(\frac{60}{-14}\)

x = \(-\frac{30}{7}\)

b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)

(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x² + 2x

(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x² + 2x

(x - 4)(-x + 8) = 5x - 10 - x² + 2x

-x² + 8x + 4x - 32 = 5x - 10 - x² + 2x

(-x² + x²) + (8x + 4x - 5x - 2x) = -10 + 32

5x = 22

x = \(\frac{22}{5}\) 

a) 5(3x+5)-4(2x-3)=5x+8(2x+12)+1

<=> 15x+25-8x+12=5x+16x+96+1

<=>15x-8x-5x-16x=-25-12+96+1

<=> -14x       = 60

<=>  x= \(\frac{-60}{14}\)=\(\frac{-30}{7}\)

Vậy x= \(\frac{-30}{7}\)

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

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

a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 ) 

<=> 2x - 3 - x + 5 = x + 7 - x - 2

<=> x = 3

b)(7x-5)-(6x+4)=(2x+3)-(2x+1)

<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1

<=> x = 11

c)(9x-3)-(8x+5)=(3x+2)

<=> 9x - 3 - 8x - 5 = 3x + 2

<=> -2x = 10

<=> x = -5

d)(x+7)-(2x+3)=(3x+5)-(2x+4)

<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4

<=> -2x = -3

<=> x = 3/2

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