a) \(A=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2=\left(2x-1+3x+1\right)^2=\left(5x\right)^2=25x^2\)
Thay \(x=-\frac{1}{5}\) vào A ta cọ
\(A=25\cdot\left(-\frac{1}{5}\right)^2=25\cdot\frac{1}{25}=1\)
b)\(B=\left(4x-3\right)^2-2\left(16x^2-9\right)+\left(4x+3\right)^2\)
\(=\left(4x-3\right)^2-2\left(4x-3\right)\left(4x+3\right)+\left(4x+3\right)^2=\left(4x-3-4x-3\right)^2=-6^2\)
Vói x=-17 thì \(B=\left(-6\right)^2=36\)
A=(2x-1)2+2.(2x-1)(3x-1)+(3x+1)2
A=4x2-4x+1+2.(6x2-5x+1)+9x2+6x+1
A=13x2+2x+2+12x2-10x+2
A=25x2-8x+4
A=(5x)2-2.4x+22
A=(5x-2)2 Tại x=\(-\frac{1}{5}\) ta có:
A=(5x-2)2
A=\(\left[5.\left(-\frac{1}{5}\right)-2\right]^2\)
A=\(\left(-1-1\right)^2\)
A=(-2)2=4
b)B=(4x-3)2-2.(16x2-9)+(4x+3)2
B=16x2-24x+32-32x2+18+16x2+24x+32
B=32+32+18
B=36
A= \(\left(2x-1+3x+1\right)^2\)
= \(5x^2\)=\(5.\left(\frac{1}{5}\right)^2\)=\(\frac{1}{5}\)
