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15 tháng 8 2020

a) \(4x^2-12x+9\)

\(=\left(2x\right)^2-2.2x.3+3^2\)

\(=\left(2x-3\right)^2\)

b) \(4x^2+4x+1\)

\(=\left(2x\right)^2+2.2x.1+1^2\)

\(=\left(2x+1\right)^2\)

c) \(1+12x+36x^2\)

\(=1^2+2.1.6x+\left(6x\right)^2\)

\(=\left(1+6x\right)^2\)

d) \(9x^2-24xy+16y^2\)

\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)

\(=\left(3x-4y\right)^2\)

e) \(\frac{x^2}{4}+2xy+4y^2\)

\(=\left(\frac{x}{2}\right)^2+2.\frac{x}{2}.2y+\left(2y\right)^2\)

\(=\left(\frac{x}{2}+2y\right)^2\)

f) \(-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x^2-2.5x+5^2\right)\)

\(=-\left(x-5\right)^2\)

g) \(-16a^4b^6-24a^5b^5-9a^6b^4\)

\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)

\(=-a^4b^4\left[\left(4b\right)^2+2.4b.3a+\left(3a\right)^2\right]\)

\(=-a^4b^4\left(4b+3a\right)^2\)

h) \(25x^2-20xy+4y^2\)

\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)

\(=\left(5x-2y\right)^2\)

i) \(25x^4-10x^2y+y^2\)

\(=\left(5x^2\right)^2-2.5x^2y+y^2\)

\(=\left(5x^2-y\right)^2\)

16 tháng 8 2018

a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)

b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)

c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)

d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)

f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)

                             \(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)

                              \(=-\left(4a^2b^3+3a^3b^2\right)^2\)

h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)

i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)

30 tháng 7 2020

a) \(4x^2-12x+9\)

\(=\left(2x\right)^2-2.2.3+3^2\)

\(=\left(2x-3\right)^2\)

b) \(4x^2+4x+1\)

\(=\left(2x\right)^2+2.2x.1+1^2\)

\(=\left(2x+1\right)^2\)

c) \(1+12x+36x^2\)

\(=1^2+2.6x+\left(6x\right)^2\)

\(=\left(1+6x\right)^2\)

d) \(9x^2-24xy+16y^2\)

\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)

\(=\left(3x-4y\right)^2\)

e) Viết = công thức trực quan hộ mình

f) \(-x^2+10x-25\)

\(=-\left(x^2-10x+25\right)\)

\(=-\left(x^2-2.5x+5^2\right)\)

\(=-\left(x-5\right)^2\)

28 tháng 9 2018

\(4x^2+4x+1\)

\(=\left(2x\right)^2+2.2x.1+1\)

\(=\left(2x+1\right)^2\)

\(1+12x+36x^2\)

\(=1+2.6x+\left(6x\right)^2\)

\(=\left(1+6x\right)^2\)

a) Ta có: \(\left(4x^2-12x+9\right)-1\)

\(=\left(2x-3\right)^2-1^2\)

\(=\left(2x-3-1\right)\left(2x-3+1\right)\)

\(=\left(2x-4\right)\left(2x-2\right)\)

\(=4\left(x-2\right)\left(x-1\right)\)

b) Ta có: \(\left(\frac{x^2}{4}+2xy+4y^2\right)-25\)

\(=\left[\left(\frac{x}{2}\right)^2+2\cdot\frac{x}{2}\cdot2y+\left(2y\right)^2\right]-5^2\)

\(=\left(\frac{x}{2}+2y\right)^2-5^2\)

\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)

c) Ta có: \(1+12x+35x^2\)

\(=35x^2+12x+1\)

\(=35x^2+5x+7x+1\)

\(=5x\left(7x+1\right)+\left(7x+1\right)\)

\(=\left(7x+1\right)\left(5x+1\right)\)

d) Ta có: \(9x^2-24xy+15y^2\)

\(=9x^2-9xy-15xy+15y^2\)

\(=9x\left(x-y\right)-15y\left(x-y\right)\)

\(=\left(x-y\right)\left(9x-15y\right)\)

\(=3\left(x-y\right)\left(3x-5y\right)\)

e) Ta có: \(25x^2-20xy+3y^2\)

\(=25x^2-15xy-5xy+3y^2\)

\(=5x\left(5x-3y\right)-y\left(5x-3y\right)\)

\(=\left(5x-3y\right)\left(5x-y\right)\)

f) Ta có: \(24x^4-10x^2y+y^2\)

\(=24x^4-4x^2y-6x^2y+y^2\)

\(=4x^2\left(6x^2-y\right)-y\left(6x^2-y\right)\)

\(=\left(6x^2-y\right)\left(4x^2-y\right)\)

a: 2x^2y-50xy=2xy(x-25)

b: 5x^2-10x=5x(x-2)

c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)

d: \(x^2-xy+x=x\left(x-y+1\right)\)

e: x(x-y)-2(y-x)

=x(x-y)+2(x-y)

=(x-y)(x+2)

f: 4x^2-4xy-8y^2

=4(x^2-xy-2y^2)

=4(x^2-2xy+xy-2y^2)

=4[x(x-2y)+y(x-2y)]

=4(x-2y)(x+y)

f1: x^2ỹ-y^2+y

=(x-y)(x+y)+(x+y)

=(x+y)(x-y+1)

26 tháng 12 2021

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26 tháng 12 2021

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)