câu 1;phân tích đa thức thành nhân tử
a, 5x^3y – 10x^2y^2 + 5xy^3
B, x^2 (2y – 1) – 125 (2y-1)
Câu 2: tìm x,biết:
a, x(x-2)-(x-3)(x+4)=0
b,(2x-1)^2-(4x^2-1)=0
c, x^2+ 2020x-2021=0
câu 1;phân tích đa thức thành nhân tử a, 5x^3y – 10x^2y^2 + 5xy^3 B, x^2 (2y – 1) – 125 (2y-1) Câu 2: tìm x,biết: a, x(x-2)-(x-3)(x+4)=0 b,(2x-1)^2-(4
By Alice
`text{Câu 1 }`
`a, 5x^3y – 10x^2y^2 + 5xy^3`
`=5xy(x^2-2xy+y^2)`
`=5xy(x-y)^2`
`b, x^2 (2y – 1) – 125 (2y-1)`
`=(x^2-125)(2y-1)`
`text{Câu 2 }`
`a, x(x-2)-(x-3)(x+4)=0`
`<=>x^2-2x-x^2-4x+3x+12=0`
`<=>-3x+12=0`
`<=>-3x=-12`
`<=>x=4`
Vậy `x` e `{4}`
`b,(2x-1)^2-(4x^2-1)=0`
`<=>(2x-1)^2-[(2x^2-1)(2x^2+1)]=0`
`<=>(2x−1)(2x−1−2x−1)=0`
`<=>-2(2x-1)=0`
`<=>2x-1=0`
`<=>x=1/2`
Vậy `x` e `{1/2}`
`c, x^2+ 2020x-2021=0`
`<=>(x-1)(x+2021)=0`
`<=>x-1=0` hoặc `x+2021=0`
`<=>x=1` hoặc `x=-2021`
Vậy `x` e `{1,-2021}`
$\begin{array}{l}\bf Câu \,\,1:\\ a)\quad 5x^3y -10x^2y^2+ 5xy^3\\ = 5xy(x^2 – 2xy + y^2)\\ = 5xy(x-y)^2\\ b)\quad x^2(2y – 1) – 125(2y-1)\\ = (2y-1)(x^2-125)\\ \bf Câu\,\,2:\\ a)\quad x(x-2) – (x-3)(x+4) = 0\\ \Leftrightarrow x^2 – 2x – (x^2 +x – 12) = 0\\ \Leftrightarrow x^2 – 2x – x^2 – x + 12 =0\\ \Leftrightarrow -3x + 12 =0\\ \Leftrightarrow 3x = 12\\ \Leftrightarrow x = 4\\ b)\quad (2x-1)^2 – (4x^2 – 1) = 0\\ \Leftrightarrow (2x-1)^2 – (2x-1)(2x+1) =0\\ \Leftrightarrow (2x-1)[(2x-1) – (2x+1)] =0\\ \Leftrightarrow (2x-1)(-2) =0\\ \Leftrightarrow 2x – 1 =0\\ \Leftrightarrow x = \dfrac12\\ c)\quad x^2 + 2020x – 2021 =0\\ \Leftrightarrow x^2 + 2021x – x – 2021 =0\\ \Leftrightarrow x(x+2021) – (x+2021) =0\\ \Leftrightarrow (x+2021)(x-1) =0\\ \Leftrightarrow \left[\begin{array}{l}x + 2021 =0\\x – 1 =0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -2021\\x = 1\end{array}\right. \end{array}$