Question

Mọi người giúp tớ câu này với ạ! 

A=\(3\left(x-3\right)\left(x-7\right)+\left(x+4\right)^2+48\) tại \(x=0,5\)
Mình cảm ơn!

Answer 1

\(A=3\left(x-3\right)\left(x+7\right)+\left(x+4\right)^2+48\)

\(A=3\left(x^2-4x-21\right)+\left(x^2+8x+16\right)+48\)

\(A=\left(3x^2+x^2\right)-\left(12x-8x\right)-\left(21-16-48\right)\)

\(A=4x^2-4x+43\)

\(A=\left(4x^2-4x+1\right)+42\)

\(A=\left(2x+1\right)^2+42\)

Thay \(x=\frac{1}{2}\) vao A ta duoc:

\(A=\left(2\cdot\frac{1}{2}+1\right)^2+42=46\)

Answer 2

\(A=3\left(x-3\right)\left(x-7\right)+\left(x+4\right)^2+48\)

\(=3x^2-13x+63+x^2+8x+16+48\)

\(=4x^2-5x+127\)

\(4\cdot0,25-5\cdot0,5+127=1-1+127=127\)