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Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
a)\(=\left(x-2\right)\left(5x-3x^2\right)\)
\(=x\left(x-2\right)\left(5-3x\right)\)
b)\(=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)
\(=-\left(x^2-4x+4\right)\left(x+2\right)^2\)
=\(-\left(x-2\right)^2\left(x+2\right)^2\)
c)\(\left(=5x^2-5xy\right)+\left(7y-7x\right)\)
\(=-5x\left(y-x\right)+7\left(y-x\right)\)
\(=\left(y-x\right)\left(7-5x\right)\)
d)\(=\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
e)\(=[\left(x^2\right)^2+8^2+2.x^2.8]-2.8.x^2\)
\(=\left(x^2+8\right)^2-16x^2\)
\(=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)
__Chúc bạn hc tốt ......
Bài 1:
a) \(a^2-6a+9=\left(a-3\right)^2\)
b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)
Bài 2:
a) \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)
\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)
b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
b)x^4+5x^2-6
=x4-x3+x3-x2+6x2-6x+6x-6
=x3(x-1)+x2(x-1)+6x(x-1)+6(x-1)
=(x-1)(x3+x2+6x+6)
=(x-1)[x2(x+1)+6(x+1)]
=(x-1)(x+1)(x2+6)