Question

Me ajudem
L) (3x+2).(2x+1) =
M) (x+7).(x-4) =
N) (3x+4).(2x-1)=
O) (x-4y).(x-y) =
P) ( 10x² + 6x) : (-2x) =
Q) (4x³ - 9x) : (+3x) =
R) ( 15x³ - 10x²) : (5x²)
S) (x² + x³ + x⁴) : (+x²) =
T) (3x⁴ - 6x³ + 10x²) : (-2x²) =
U) ( x² + 5x + 6) : (x + 2) =
V) (x² - 7x + 10 ) : ( x - 2)
W) ( 4x⁴ - 14x³ + 15x² -17x + 5 ) : (x² - 3x + 1)

Answer 1

Resposta:

$L) 6x^2+7x+2\\M) x^2+3x-28\\N) 6x^2+5x-4\\O) x^2-5xy+4y^2\\P) -5x-3\\Q) \frac{4x^2-9}{3}\\R) 3x-2\\S) x^2+x+1\\T) -\frac{3x^2-6x+10}{2}\\U) x+3\\V) x-5\\W) 4x^2-2x+5$

Explicação passo-a-passo:

$L) \left(3x+2\right)\left(2x+1\right)\\=3x\cdot \:2x+3x\cdot \:1+2\cdot \:2x+2\cdot \:1\\=3\cdot \:2xx+3\cdot \:1\cdot \:x+2\cdot \:2x+2\cdot \:1\\=6x^2+7x+2$

$M) \left(x+7\right)\left(x-4\right)\\=xx+x\left(-4\right)+7x+7\left(-4\right)\\=xx-4x+7x-7\cdot \:4\\=x^2+3x-28$

$N) \left(3x+4\right)\left(2x-1\right)\\=3x\cdot \:2x+3x\left(-1\right)+4\cdot \:2x+4\left(-1\right)\\=3\cdot \:2xx-3\cdot \:1\cdot \:x+4\cdot \:2x-4\cdot \:1\\=6x^2+5x-4$

$O) \left(x-4y\right)\left(x-y\right)\\=xx+x\left(-y\right)+\left(-4y\right)x+\left(-4y\right)\left(-y\right)\\=xx-xy-4xy+4yy\\=x^2-5xy+4y^2$

$P) \frac{10x^2+6x}{-2x}\\=-\frac{10x^2+6x}{2x}\\=-\left(5x+3\right)\\=-\left(5x\right)-\left(3\right)\\=-5x-3$

$Q) \frac{4x^3-9x}{3x}\\=\frac{x\left(4x^2-9\right)}{3x}\\=\frac{4x^2-9}{3}$

$R) \frac{15x^3-10x^2}{5x^2}\\=\frac{5x^2\left(3x-2\right)}{5x^2}\\=3x-2$

$S) \frac{x^2+x^3+x^4}{x^2}\\=1+x+x^2\\=x^2+x+1$

$T) \frac{3x^4-6x^3+10x^2}{-2x^2}\\=-\frac{3x^4-6x^3+10x^2}{2x^2}\\=-\frac{3x^2-6x+10}{2}$

$U) \frac{x^2+5x+6}{x+2}\\=\frac{\left(x+2\right)\left(x+3\right)}{x+2}\\=x+3$

$V) \frac{x^2-7x+10}{x-2}\\=\frac{\left(x-2\right)\left(x-5\right)}{x-2}\\=x-5$

$W) \frac{\left(4x^4-14x^3+15x^2-17x+5\right)}{\left(x^2-3x+1\right)}\\=4x^2+\frac{-2x^3+11x^2-17x+5}{x^2-3x+1}\\=4x^2-2x+\frac{5x^2-15x+5}{x^2-3x+1}\\=4x^2-2x+5$