Lời giải:
a.
$A=x^2+6x+9+(x^2-9)-2(x^2-2x-8)$
$=10x+16=10.\frac{-1}{2}+16=-5+16=11$
b.
$B=9x^2+24x+16-(x^2-16)-10x$
$=8x^2+14x+32=8(\frac{-1}{10})^2+14.\frac{-1}{10}+32$
$=\frac{767}{25}$
c.
$C=(x^2+2x+1)-(4x^2-4x+1)+3(x^2-4)$
$=6x-12=6-12=-6$
d.
$D=(x^2-9)+(x^2-4x+4)-(2x^2-8x)$
$=4x-5=4(-1)-5=-9$
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