Phân tích đa thức thành nhân tử: a,5(x-y)-(x-y) ; b,x^2-y^2+5x-5y ; c, 1- 8x+16x^2-y^2 ; d, 5x^2y-35xy+60y 28/08/2021 Bởi Katherine Phân tích đa thức thành nhân tử: a,5(x-y)-(x-y) ; b,x^2-y^2+5x-5y ; c, 1- 8x+16x^2-y^2 ; d, 5x^2y-35xy+60y
Giải thích các bước giải: $\begin{array}{l} a,\,5\left( {x – y} \right) – \left( {x – y} \right)\\ = \left( {x – y} \right)\left( {5 – 1} \right)\\ = 4\left( {x – y} \right)\\ b,\,{x^2} – {y^2} + 5x – 5y\\ = \left( {x – y} \right)\left( {x + y} \right) + 5\left( {x – y} \right)\\ = \left( {x – y} \right)\left( {x + y + 5} \right)\\ c,\,1 – 8x + 16{x^2} – {y^2}\\ = {(4x)^2} – 2.4x.1 + 1 – {y^2}\\ = {(4x – 1)^2} – {y^2}\\ = (4x – 1 – y)(4x – 1 + y)\\ d,\,5{x^2}y – 35xy + 60y\\ = 5y({x^2} – 7x + 12)\\ = 5y({x^2} – 3x – 4x + 12)\\ = 5y\left[ {x(x – 3) – 4(x – 3)} \right]\\ = 5y(x – 3)(x – 4) \end{array}$ Bình luận
a, $5(x-y)-(x-y)$ $=(x-y)(5-1)$ $=4(x-y)$ b, $x^2-y^2+5x-5y$ $=(x-y)(x+y)+5(x-y)$ $=(x-y)(x+y+5)$ c, $1-8x+16x^2-y^2$ $=1^2-2.4.x+(4x)^2-y^2$ $=(1-4x)^2-y^2$ $=(1-4x-y)(1-4x+y)$ d, $5x^2y-35xy+60y$ $=5y(x^2-7x+12)$ $=5y(x^2-3x-4x+12)$ $=5y[x(x-3)-4(x-3)]$ $=5y(x-3)(x-4)$ Bình luận
Giải thích các bước giải:
$\begin{array}{l} a,\,5\left( {x – y} \right) – \left( {x – y} \right)\\ = \left( {x – y} \right)\left( {5 – 1} \right)\\ = 4\left( {x – y} \right)\\ b,\,{x^2} – {y^2} + 5x – 5y\\ = \left( {x – y} \right)\left( {x + y} \right) + 5\left( {x – y} \right)\\ = \left( {x – y} \right)\left( {x + y + 5} \right)\\ c,\,1 – 8x + 16{x^2} – {y^2}\\ = {(4x)^2} – 2.4x.1 + 1 – {y^2}\\ = {(4x – 1)^2} – {y^2}\\ = (4x – 1 – y)(4x – 1 + y)\\ d,\,5{x^2}y – 35xy + 60y\\ = 5y({x^2} – 7x + 12)\\ = 5y({x^2} – 3x – 4x + 12)\\ = 5y\left[ {x(x – 3) – 4(x – 3)} \right]\\ = 5y(x – 3)(x – 4) \end{array}$
a, $5(x-y)-(x-y)$
$=(x-y)(5-1)$
$=4(x-y)$
b, $x^2-y^2+5x-5y$
$=(x-y)(x+y)+5(x-y)$
$=(x-y)(x+y+5)$
c, $1-8x+16x^2-y^2$
$=1^2-2.4.x+(4x)^2-y^2$
$=(1-4x)^2-y^2$
$=(1-4x-y)(1-4x+y)$
d, $5x^2y-35xy+60y$
$=5y(x^2-7x+12)$
$=5y(x^2-3x-4x+12)$
$=5y[x(x-3)-4(x-3)]$
$=5y(x-3)(x-4)$