Câu 1.2: Rút gọn biểu thức:
a) \(2x\left(x+3\right)-x\left(2x-1\right)\)
\(=2x^2+6x-2x^2+x\)
\(=7x\)
b) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
\(=\left(x-2y\right)\left(x^2+2xy+\left(2y\right)^2\right)\)
\(=x^3-\left(2y\right)^3\)
\(=x^3-8y^3\)
c) \(\left(x+2\right)^2+\left(x-2\right)^2\)
\(=x^2+4x+4+x^2-4x+4\)
\(=2x^2+8\)
\(=2\left(x^2+4\right)\)
d) \(\left(4y+1\right)^2-4\left(2y+3\right)\left(2y-3\right)\)
\(=\left(4y+1\right)^2-4\left(\left(2y\right)^2-3^2\right)\)
\(=\left(4y+1\right)^2-4\left(4y^2-9\right)\)
\(=16y^2+8y+1-16y^2+36\)
\(=8y+37\)
Câu 2: Tìm x:
b) \(10x+25+x^2=0\)
<=> \(x^2+2.5x+5^2=0\)
<=> \(\left(x+5\right)^2=0\)
=> \(x+5=0\)
<=> \(x=-5\)
Vậy x= \(-5\)
a) \(2\left(x^2+3\right)-2x\left(x+3\right)=0\)
<=> \(2x^2+6-2x^2-6x=0\)
<=> \(6-6x=0\)
<=> \(6\left(1-x\right)=0\)
=> \(1-x=0\)
<=> \(x=1\)
Vậy x = 1
