Bonsoir Merci de m'aider a faire ces deux exercices.

Bonjour Bibigarnier

Factoriser :

[tex]1)\ (x+1)(3-2x)+(3-2x)^2\\=(x+1)(3-2x)+(3-2x)(3-2x)\\=(3-2x)[(x+1)+(3-2x)]\\=(3-2x)(x+1+3-2x)\\=(3-2x)(-x+4)\\\\\boxed{ (x+1)(3-2x)+(3-2x)^2=(3-2x)(-x+4)}[/tex]

[tex]2)\ (3x+4)(2x-1)+4(3x+4)\\=(3x+4)[(2x-1)+4]\\=(3x+4)(2x-1+4)\\=(3x+4)(2x+3)\\\\\boxed{(3x+4)(2x-1)+4(3x+4)=(3x+4)(2x+3)}[/tex]

[tex]3)\ x(2-x)+(3x+1)(2-x)\\=(2-x)[x+(3x+1)]\\=(2-x)(4x+1)\\\\\boxed{x(2-x)+(3x+1)(2-x)=(2-x)(4x+1)}[/tex]

Mettre sous le même dénominateur :

[tex]1)\dfrac{2x-2}{x-1}+\dfrac{3x+3}{2x+1}=\dfrac{(2x-2)(2x+1)}{(x-1)(2x+1)}+\dfrac{(3x+3)(x-1)}{(2x+1)(x-1)}\\\\=\dfrac{(2x-2)(2x+1)+(3x+3)(x-1)}{(x-1)(2x+1)}\\\\=\dfrac{2(x-1)(2x+1)+(3x+3)(x-1)}{(x-1)(2x+1)}\\\\=\dfrac{(x-1)[2(2x+1)+(3x+3)]}{(x-1)(2x+1)}\\\\=\dfrac{(x-1)(4x+2+3x+3)}{(x-1)(2x+1)}[/tex]

[tex]\\\\=\dfrac{(x-1)(7x+5)}{(x-1)(2x+1)}\\\\\\\boxed{ \dfrac{2x-2}{x-1}+\dfrac{3x+3}{2x+1}=\dfrac{(x-1)(7x+5)}{(x-1)(2x+1)}}\\\\\\\boxed{ \dfrac{2x-2}{x-1}+\dfrac{3x+3}{2x+1}=\dfrac{7x+5}{2x+1}\ \ si\ x\neq1}[/tex]

[tex]2)\ 1+\dfrac{2x+1}{x}\\\\=\dfrac{x}{x}+\dfrac{2x+1}{x}\\\\=\dfrac{x+(2x+1)}{x}\\\\=\dfrac{x+2x+1}{x}\\\\=\dfrac{3x+1}{x}\\\\\\\boxed{1+\dfrac{2x+1}{x}=\dfrac{3x+1}{x}}[/tex]

[tex]3)\ \dfrac{(2x-3)^2}{x}-\dfrac{4}{x+1}\\\\=\dfrac{(2x-3)^2(x+1)}{x(x+1)}-\dfrac{4x}{x(x+1)}\\\\=\dfrac{(2x-3)^2(x+1)-4x}{x(x+1)}\\\\=\dfrac{(4x^2-12x+9)(x+1)-4x}{x(x+1)}\\\\=\dfrac{(4x^3+4x^2-12x^2-12x+9x+9)-4x}{x(x+1)}\\\\=\dfrac{4x^3-8x^2-7x+9}{x(x+1)}[/tex]

[tex]\\\\\boxed{\dfrac{(2x-3)^2}{x}-\dfrac{4}{x+1}=\dfrac{4x^3-8x^2-7x+9}{x(x+1)}}[/tex]