Question

Tính nhanh giá trị biểu thức :

a) x²+4y²-4xy tại x = 18, y = 4

b) (2x+1)²+(2x-1)²-2(1+2x)(1-2x) tại x = 100

Answer 1

a) \(x^2-4xy+4y^2\)

\(=x^2-2.x.2y+\left(2y\right)^2\)

\(=\left(x-2y\right)^2\)

Thay x = 18 ; y = 4 vào ta được

\(=\left(18-2.4\right)^2\)

\(=10^2=100\)

b) \(\left(2x+1\right)^2-2\left(1+2x\right)\left(1-2x\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)

\(=\left(2x+1+2x-1\right)^2\)

\(=\left(4x\right)^2\)

Thay x = 100 ta được

\(=\left(4.100\right)^2\)

\(=400^2=160000\)

Answer 2

a) ta có : \(x^2+4y^2-4xy=\left(x-2y\right)^2=\left(18-2.4\right)^2=100\)

b) ta có : \(\left(2x+1\right)^2+\left(2x-1\right)^2-2\left(1+2x\right)\left(1-2x\right)\)

\(=\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(2x+1\right)\left(2x-1\right)\)

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2=16\left(100\right)^2=160000\)