a: \(A=x^2-8x+7\)
\(=x^2-x-7x+7\)
\(=x\left(x-1\right)-7\left(x-1\right)\)
\(=\left(x-1\right)\cdot\left(x-7\right)\)
Thay x=51 vào A, ta được:
\(A=\left(51-1\right)\left(51-7\right)=50\cdot44=2200\)
b: \(B=x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
Thay x=10 vào B, ta được:
\(B=10^2\left(10-1\right)^2=9^2\cdot100=8100\)
c: \(C=x^2+4xy+4y^2-z^2\)
\(=\left(x^2+4xy+4y^2\right)-z^2\)
\(=\left(x+2y\right)^2-z^2\)
\(=\left(x+2y+z\right)\left(x+2y-z\right)\)
Khi x=5 và y=7 và z=19 thì \(C=\left(5+2\cdot7+19\right)\left(5+2\cdot7-19\right)=0\)
d:
\(m^3-m+1=0\)
=>\(m^3=m-1\)
\(D=m^6-2m^4-m+m^2+m^3\)
\(=\left(m^6-m^4+m^3\right)+\left(m^4+m^2-m\right)\)
\(=m^3\left(m^3-m+1\right)+m\left(m^3+m-1\right)\)
\(=m^3\cdot0+m\left(m-1+m-1\right)\)
\(=0+m\cdot2m=2m^2\)
Lời giải:
a. $A=x^2-8x+7=(x-1)(x-7)=(51-1)(51-7)=50.44=2200$
b. $B=x^4-2x^3+x^2=(x^2-x)^2=(10^2-10)^2=90^2=8100$
c. $C=(x+2y)^2-z^2=(x+2y-z)(x+2y+z)$
$=(5+14-19).(5+14+19)=0(5+14+19)=0$
d.
$D=(m^6-2m^4+m^2)+(m^3-m)$
$=(m^3-m)^2+(m^3-m)=(m^3-m)(m^3-m+1)=(m^3-m).0=0$
