a: \(A=4x^2-12xy+9y^2\)
\(=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2\)
\(=\left(2x-3y\right)^2\)
Thay x=2;y=-1 vào A, ta được:
\(A=\left[2\cdot2-3\cdot\left(-1\right)\right]^2=\left(4+3\right)^2=7^2=49\)
b: \(B=5\left(x-3\right)\left(x+3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)
\(=5\left(x^2-9\right)+4x^2+12x+9+x^2-12x+36\)
\(=5x^2-45+5x^2+45=10x^2\)
Khi x=1/5 thì \(B=10\cdot\left(\dfrac{1}{5}\right)^2=\dfrac{10}{25}=\dfrac{2}{5}\)
`A = 4x^2 - 12xy + 9y^2`
`=> A = (2x)^2 - 2.2x.3y + (3y)^2 `
`=> A = (2x - 3y)^2 (hằng đẳng thức số 2)
thay `x = 2 ; y = -1` vào `A` ta có :
`A = [2.2 - 3.(-1)]^2`
`=> A = (4+3)^2`
`=> A = 7^2 = 49`
Vậy `A = 49 ` khi `x = 2 ; y = -1`
`B = 5(x-3)(x+3) + (2x + 3)^2 + (x-6)^2`
`=> B = 5(x^2 - 9) + (4x^2 + 12x + 9) + (x^2 - 12x + 36)`
`=> B = 5x^2 - 45 + 4x^2 + 12x + 9 + x^2 - 12x + 36`
`=> B = (5x^2 + 4x^2 + x^2) + (12x - 12x) - (45 - 9 - 36)`
`=> B = 10x^2 `
Thay `x = 1/5` vào `B` ta có :
`B = 10.(1/5)^2`
`=> B = 10 .1/25`
`=> B = 2/5`
Vậy `B = 2/5` với `x = 1/5`