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a) \(36-4x^2+4xy-y^2\)
\(=36-\left(2x-y\right)^2\)
\(=\left(6+2x-y\right)\left(6-2x+y\right)\)
b) \(2x^4+3x^2-5\)
\(=2x^4-2x^2+5x^2-5\)
\(=2x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(2x^2+5\right)\left(x+1\right)\left(x-1\right)\)
\(a,2x^2+3x+1\)
\(=2x^2+2x+x+1\)
\(=\left(2x^2+2x\right)+\left(x+1\right)\)
\(=2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+1\right)\)
\(b,2x^2+6x+3\)
\(=2x^2+2x+3x+3\)
\(=2x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+3\right)\)
a,\(x^2+4x+3\)
=\(x^2+3x+x+3\)
=\(x\left(x+3\right)+\left(x+3\right)\)
=(x+3)(x+1)
b,\(2x^2+3x-5\)
=\(2x^2+5x-2x-5\)
=x(2x+5)-(2x+5)
=(2x+5)(x-1)
c,\(16x-5x^2-3\)
=\(-\left(5x^2-16x+3\right)\)
=\(-\left(5x^2-x-15x+3\right)\)
=-[x(5x-1)-3(5x-1)]
=-[(5x-1)(x-3)]
=-(5x-1)(x-3)
\(a.\) \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+3\right)\left(x+1\right)\)
\(b.\) \(2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(c.\)\(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(1-5x\right)\)
d) \(2x^2-3x-27\)
\(=\left(2x^2+6x\right)-\left(9x+27\right)\)
\(=2x\left(x+3\right)-9\left(x+3\right)\)
\(=\left(2x-9\right)\left(x+3\right)\)
e) \(2x^2-5xy-3y^2\)
\(=\left(2x^2+xy\right)-\left(6xy+3y^2\right)\)
\(=2x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(2x-3y\right)\left(x+y\right)\)
a) \(x^3-3x+1-3x^2=\left(x^3+1\right)-\left(3x^2+3x\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)=\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(2x^2+4x+2-2y^2=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x+1+y\right)\left(x+1-y\right)\)
Ta có:
\(2x^2-5xy+3y^2\)
\(=2x^2-2xy-3xy+3y^2=2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-3y\right)\)
\(x^3-7x-6=x^3+1-7x-7\)
\(=\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
\(2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5 \left(x-1\right)=\left(x-1\right)\left(2x+5\right) \)
bảng 1