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Phân tích thành nhân tử : -4x^2+12xy-9y^2+25(sử dụng hằng đẳng thức)
giúp mink với, mink đang cần gấp
\(=-2x^2+6xy-3y^2+25\)
\(=-\left(2x^2-6xy+3y^2\right)+25\)
\(=-\left(2x^2+3y^2\right)+25\)
\(=\left(2x^2-3y^2\right)+25\)
\(=\left(2x^2+3y^2\right).\left(2x^2-3y^2\right)+25\)
a/25-9y^2-4x^2+12xy
=-(9y^2-12xy+4x^2-25)
=-[(3y)^2-2.3y.2x+(2x)^2-5^2]
=-[(3y-2x)^2-5^2]
=-(3y-2x-5)(3y-2x+5)
b/4x^2-8xy+4y^2-5x+5y
=4(x^2-2.x.y+y^2)-5(x-y)
=4(x-y)^2-5(x-y)
=(x-y)(4x-4y-5)
a: \(ab+a+b+1\)
\(=a\left(b+1\right)+\left(b+1\right)\)
\(=\left(b+1\right)\left(a+1\right)\)
c: \(4x^2-12xy+3x-9y\)
\(=4x\left(x-3y\right)+3\left(x-3y\right)\)
\(=\left(x-3y\right)\left(4x+3\right)\)
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
mình k ghi lại đề nha bạn
\(=\left(x-y\right)^2-16z^2\\ =\left(x-y-4z\right)\left(x-y+4z\right)\)
\(25-4x^2+12xy-9y^2\)
\(=25-\left(2x-3y\right)^2\)
\(=\left(5-2x+3y\right)\left(5+2x-3y\right)\)
\(-4x^2+12xy-9y^2+25=25-\left(2x-3y\right)^2=\left(5-2x+3y\right)\left(5+2x-3y\right)\)
\(-4x^2+12xy-9y^2+25\)
\(=25-\left(4x^2-12xy+9y^2\right)\)
\(=25-\left(2x-3\right)^2\)
\(=\left(5+2x-3\right)\left(5-2x+3\right)\)