Question

Tính:

\(A=\left(x-10\right)^2-x\left(x+80\right)\)tại \(x=0,98\)

\(B=4x^2-28x+49\)tại \(x=4\)

\(C=x^3-9x^2+27x-27\) tại \(x=5\)

Answer 1

\(A=\left(0,98-10\right)^2-0,98\)\(\left(0,98+80\right)\)\(\Rightarrow A=0,98^2-2\times0,98\times10+10^2-0,98^2-0,98\times80\)

\(\Rightarrow A=\left(0,98^2-0,98^2\right)-\left(2\times0,98\times10+0,98\times80\right)+10^2\)

\(\Rightarrow A=0-\left(0,98\left(2\times10+80\right)\right)+100\)

\(\Rightarrow A=0-\left(0,98\times100\right)+100\Rightarrow A=0-98+100\Rightarrow A=2\)

Answer 2

b)\(B=4\times4^2-28\times4+49\Rightarrow B=4\left(4^2-28\right)+49\Rightarrow B=4\times\left(-12\right)+49\)\(\Rightarrow B=-48+49\Rightarrow B=1\)

c)\(C=5^3-9\times5^2+27\times5-27\Rightarrow C=5^2\left(5-9\right)+27\left(5-1\right)\Rightarrow C=25\times\left(-4\right)+27\times4\)

\(\Rightarrow C=4\left(-25+27\right)\Rightarrow C=4\times2\Rightarrow C=8\)