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1: \(=\left(3x-2y\right)^2-3\)
2: \(=x^2+4x+4-3=\left(x+2\right)^2-3\)
3: \(=x^2-4x+4+3=\left(x-2\right)^2+3\)
5 \(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6: \(=\dfrac{1}{4}x^2+x+1-1=\left(\dfrac{1}{2}x+1\right)^2-1\)
Giải:
1) \(9x^2-12xy+4y^2-3\)
\(=\left(9x^2-12xy+4y^2\right)-3\)
\(=\left(3x-2y\right)^2-3\)
2) \(x^2+4x+1\)
\(=x^2+4x+4-3\)
\(=\left(x+2\right)^2-3\)
3) \(x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\)
4) \(x^2+6x+15\)
\(=x^2+6x+9+6\)
\(=\left(x+3\right)^2+6\)
5) \(x^2-x+\dfrac{1}{3}\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{1}{12}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{12}\)
6) \(\dfrac{1}{4}x^2+x\)
\(=x\left(\dfrac{1}{4}x+1\right)\)
7) \(3x^2+2x+1\)
\(=x^2+2x+1+2x^2\)
\(=\left(x+1\right)^2+2x^2\)
8) \(2x^2-2x+1\)
\(=x^2-2x+1+x^2\)
\(=\left(x-1\right)^2+x^2\)
9) \(10a^2+5b^2+12ab+4a-6b+15\)
\(=a^2+b^2+9a^2+12ab+4b^2+4a-6b+15\)
\(=9a^2+12ab+4b^2+a^2+4a-6b+b^2+15\)
\(=\left(3a+2b\right)^2+a\left(a+4\right)-b\left(6-b\right)+15\)
Vậy ...