a: \(\left|x\right|=\frac12\)
=>\(\left[\begin{array}{l}x=\frac12\\ x=-\frac12\end{array}\right.\)
Thay x=1/2 vào A, ta được:
\(A=2\cdot\left(\frac12\right)^2-3\cdot\frac12+5\)
\(=2\cdot\frac14-\frac32+5=\frac12-\frac32+5=5-\frac22=5-1=4\)
Thay x=-1/2 vào A, ta được:
\(A=2\cdot\left(-\frac12\right)^2-3\cdot\left(-\frac12\right)+5\)
\(=2\cdot\frac14+\frac32+5=\frac12+\frac32+5=5+2=7\)
b: \(B=2x^2-3xy+y^2\)
\(=2x^2-2xy-xy+y^2\)
=2x(x-y)-y(x-y)
=(x-y)(2x-y)
|y|=1
=>y=1 hoặc y=-1
TH1: x=1/2; y=1
B=(x-y)(2x-y)
\(=\left(\frac12-1\right)\left(2\cdot\frac12-1\right)\)
=0
TH2: x=1/2; y=-1
B=(x-y)(2x-y)
\(=\left(\frac12+1\right)\left(2\cdot\frac12+1\right)=\frac32\cdot\left(1+1\right)=\frac32\cdot2=3\)
TH3: x=-1/2; y=1
B=(x-y)(2x-y)
\(=\left(-\frac12-1\right)\left(2\cdot\frac{-1}{2}-1\right)\)
\(=-\frac32\left(-1-1\right)=-\frac32\cdot\left(-2\right)=3\)
TH4: x=-1/2; y=-1
B=(x-y)(2x-y)
\(=\left(-\frac12+1\right)\left(2\cdot\frac{-1}{2}+1\right)=0\)
