Câu hỏi của Minh Anh

Question

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Nguyễn Lê Phước Thịnh · Accepted Answer

a: $9x^2+12x+4=\left(3x+2\right)^2$b: $x^2-10x+25=\left(x-5\right)^2$c: $4x^2+4x+1=\left(2x+1\right)^2$d: $x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2$e: $x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2$f: $4x^2y^4-12xy^2+9=\left(2xy^2-3\right)^2$g: $16x^2+24x+9=\left(4x+3\right)^2$h: $\dfrac{x^2}{4}-3x+9=\left(\dfrac{1}{2}x-3\right)^2$i: $\dfrac{25}{x^2}-\dfrac{10}{x}+1=\left(\dfrac{5}{x}-1\right)^2$Câu trả lời: a: $9x^2+12x+4=\left(3x+2\right)^2$b: $x^2-10x+25=\left(x-5\right)^2$c: $4x^2+4x+1=\left(2x+1\right)^2$d: $x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2$e: $x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2$f: $4x^2y^4-12xy^2+9=\left(2xy^2-3\right)^2$g: $16x^2+24x+9=\left(4x+3\right)^2$h: $\dfrac{x^2}{4}-3x+9=\left(\dfrac{1}{2}x-3\right)^2$i: $\dfrac{25}{x^2}-\dfrac{10}{x}+1=\left(\dfrac{5}{x}-1\right)^2$

ILoveMath · Answer

a) $9x^2+12x+4=\left(3x+2\right)^2$b) $x^2+25-10x=\left(x-5\right)^2$c) $4x^2+4x+1=\left(2x+1\right)^2$d) $x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2$e) $x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2$f) $4x^2y^4-12xy^2+9=\left(2xy^2-3\right)^2$g) $16x^2+24x+9=\left(4x+3\right)^2$h) $\dfrac{x^2}{4}-3x+9=\left(\dfrac{x}{2}-3\right)^2$i) $\dfrac{25}{x^2}-\dfrac{10}{x}+1=\left(\dfrac{5}{x}-1\right)^2$Câu trả lời: a) $9x^2+12x+4=\left(3x+2\right)^2$b) $x^2+25-10x=\left(x-5\right)^2$c) $4x^2+4x+1=\left(2x+1\right)^2$d) $x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2$e) $x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2$f) $4x^2y^4-12xy^2+9=\left(2xy^2-3\right)^2$g) $16x^2+24x+9=\left(4x+3\right)^2$h) $\dfrac{x^2}{4}-3x+9=\left(\dfrac{x}{2}-3\right)^2$i) $\dfrac{25}{x^2}-\dfrac{10}{x}+1=\left(\dfrac{5}{x}-1\right)^2$

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