câu 2 phân tích đa thức thành nhân tử a,11x+11y-x^2-xy b,5x^2-5y^2-7x+7y c,x^2-xy+x-y d,x^2+x-y^2+y 13/08/2021 Bởi Everleigh câu 2 phân tích đa thức thành nhân tử a,11x+11y-x^2-xy b,5x^2-5y^2-7x+7y c,x^2-xy+x-y d,x^2+x-y^2+y
Đáp án: a) 11x + 11y – x² -xy = 11(x+y)-x(x+y) = (x+y)(11-x) b) 5x² – 5y² – 7x+7y = 5(x² -y²)-7(x-y) = 5(x-y)(x+y) – 7(x-y) = (x-y){5(x+y)-7} = (x-y)(5x+5y-7) c) x² – xy + x-y = x(x-y)+(x-y) = (x-y)(x+1) d) x² + x -y² +y = (x²-y²)+(x+y) = (x-y)(x+y)+(x+y) = (x+y)(x-y+1) Bình luận
Đáp án: $\begin{array}{l}a)11x + 11y – {x^2} – xy\\ = 11\left( {x + y} \right) – x\left( {x + y} \right)\\ = \left( {x + y} \right)\left( {11 – x} \right)\\b)5{x^2} – 5{y^2} – 7x + 7y\\ = 5\left( {{x^2} – {y^2}} \right) – 7\left( {x – y} \right)\\ = 5\left( {x – y} \right)\left( {x + y} \right) – 7\left( {x – y} \right)\\ = \left( {x – y} \right)\left( {5\left( {x + y} \right) – 7} \right)\\ = \left( {x – y} \right)\left( {5x + 5y – 7} \right)\\c){x^2} – xy + x – y\\ = x\left( {x – y} \right) + \left( {x – y} \right)\\ = \left( {x – y} \right)\left( {x + 1} \right)\\d){x^2} + x – {y^2} + y\\ = \left( {{x^2} – {y^2}} \right) + \left( {x + y} \right)\\ = \left( {x – y} \right)\left( {x + y} \right) + \left( {x + y} \right)\\ = \left( {x + y} \right)\left( {x – y + 1} \right)\end{array}$ Bình luận
Đáp án:
a) 11x + 11y – x² -xy
= 11(x+y)-x(x+y)
= (x+y)(11-x)
b) 5x² – 5y² – 7x+7y
= 5(x² -y²)-7(x-y)
= 5(x-y)(x+y) – 7(x-y)
= (x-y){5(x+y)-7}
= (x-y)(5x+5y-7)
c) x² – xy + x-y
= x(x-y)+(x-y)
= (x-y)(x+1)
d) x² + x -y² +y
= (x²-y²)+(x+y)
= (x-y)(x+y)+(x+y)
= (x+y)(x-y+1)
Đáp án:
$\begin{array}{l}
a)11x + 11y – {x^2} – xy\\
= 11\left( {x + y} \right) – x\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {11 – x} \right)\\
b)5{x^2} – 5{y^2} – 7x + 7y\\
= 5\left( {{x^2} – {y^2}} \right) – 7\left( {x – y} \right)\\
= 5\left( {x – y} \right)\left( {x + y} \right) – 7\left( {x – y} \right)\\
= \left( {x – y} \right)\left( {5\left( {x + y} \right) – 7} \right)\\
= \left( {x – y} \right)\left( {5x + 5y – 7} \right)\\
c){x^2} – xy + x – y\\
= x\left( {x – y} \right) + \left( {x – y} \right)\\
= \left( {x – y} \right)\left( {x + 1} \right)\\
d){x^2} + x – {y^2} + y\\
= \left( {{x^2} – {y^2}} \right) + \left( {x + y} \right)\\
= \left( {x – y} \right)\left( {x + y} \right) + \left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x – y + 1} \right)
\end{array}$