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

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

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

30 tháng 7 2020

cái cuối hằng đẳng thức là xong mà bạn

30 tháng 7 2020

a) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

b) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left[3\left(x+y-1\right)\right]^2-\left[2\left(2x+3y+1\right)\right]^2\)

\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)

\(=\left(3x+3y-3+4x+6y+2\right)\left(3x+3y-3-4x-6y-2\right)\)

\(=\left(7x+9y-1\right)\left(-x-3y-5\right)\)

c) \(-4x^2+12xy-9y^2+25\)

\(=-\left(2x\right)^2+2.2x.3y-\left(3y\right)^2+5^2\)

\(=-\left[\left(2x\right)^2-2.2x.3y+\left(3y\right)^2-5^2\right]\)

\(=-\left[\left(2x-3y\right)^2-5^2\right]\)

\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)

d) \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-4m\left(m-n\right)-n^2\)

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

\(=\left(x-y-n\right)\left(x-y+n\right)-4m\left(m-n\right)\)

a) Ta có: \(\left(3x-1\right)^2-16\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

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

\(=3\left(x+1\right)\left(3x-5\right)\)

b) Ta có: \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)

\(=\left(-2x-4\right)\left(12x-4\right)\)

\(=-2\left(x+2\right)\cdot4\cdot\left(3x-1\right)\)

\(=-8\left(x+2\right)\left(3x-1\right)\)

c) Ta có: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+14\right)\left(3x-4\right)\)

d) Ta có: \(\left(3x+1\right)^2-4\left(x-2\right)^2\)

\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)

\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)

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

e) Ta có: \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)

\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)

\(=\left(4x+7\right)\left(8x+11\right)\)

f) Ta có: \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=-\left(b^2-2bc+c^2-a^2\right)\left[\left(b^2+2bc+c^2\right)-a^2\right]\)

\(=-\left[\left(b-c\right)^2-a^2\right]\cdot\left[\left(b+c\right)^2-a^2\right]\)

\(=-\left(b-c-a\right)\left(b-c+a\right)\left(b+c-a\right)\left(b+c+a\right)\)

g) Ta có: \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)

\(=\left[a\left(x-y\right)+b\left(y-x\right)\right]\left[a\left(x+y\right)+b\left(x+y\right)\right]\)

\(=\left[a\left(x-y\right)-b\left(x-y\right)\right]\left(x+y\right)\left(a+b\right)\)

\(=\left(x-y\right)\left(a-b\right)\left(x+y\right)\left(a+b\right)\)

h) Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)

\(=\left[\left(a^2+2ab+b^2\right)-1\right]\left[\left(a^2-2ab+b^2\right)-9\right]\)

\(=\left(a+b-1\right)\left(a+b+1\right)\left(a-b-3\right)\left(a-b+3\right)\)

i) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(x^2-9\right)\)

\(=-12\left(x+3\right)^2\cdot\left(x-3\right)\)

k) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)

\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

l) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-5^2\right]\)

\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)

m) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x-y\right)^2-\left(4m^2-4mn+n^2\right)\)

\(=\left(x-y\right)^2-\left(2m-n\right)^2\)

\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)

5 tháng 9 2017

dễ mà tự suy nghĩ và dùng máy tính bấm là ra thôi

18 tháng 8 2020

cảm ơn bạn nha

a) Ta có: 10(x-y)-8y(y-x)

\(=10\left(x-y\right)+8y\left(x-y\right)\)

\(=2\left(x-y\right)\left(5+4y\right)\)

d) Ta có: \(x^2y-x^3-9y+9x\)

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(y-x\right)\left(x^2-9\right)\)

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

e) Ta có: \(2x+2y-x^2-xy\)

\(=2\left(x+y\right)-x\left(x+y\right)\)

\(=\left(x+y\right)\left(2-x\right)\)

f) Ta có: \(x^2-25+y^2+2xy\)

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

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

g) Ta có: \(x^2-2x-4y^2-4y\)

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

h) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-16\right)\)

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

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

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

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

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

19 tháng 8 2020

a, x6 - x4 + 2x3 + 2x

= x(x5- x3 + 2x2 + 2)
b, x2 + 2x + 1 - y2

= (x + 1)2 - y2

= (x + 1 + y)(x + 1 - y)
c, x2 + 2xy + y2 - 9z2

= (x + y)2 - 9z2

= (x + y + 3z)(x + y - 3z)s
d, x3 - 10x2 + 25x - 16xy2

= x(x2 - 10x + 25 - 16y2)

= x[(x - 5)2 - 16y2]

= x(x - 5 - 4y)(x - 5 + 4y)
e, 3xy2 - 2xy + 12x

= x(3y2 - 2y + 12)

12 tháng 7 2017

       x2-4x+4=4x2-12x+9

\(\Leftrightarrow\)3x2-8x+5=0

\(\Leftrightarrow\)3x2-3x-5x+5=0

\(\Leftrightarrow\)3x(x-1)-5(x-1)=0

\(\Leftrightarrow\)(x-1)(3x-5)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=1\end{cases}}\)

b,x2-2x-25=0

\(\Leftrightarrow\)(x-1)2-26=0

\(\Leftrightarrow\)(x-1-\(\sqrt{26}\))(x-1+\(\sqrt{26}\))=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{26}+1\\x=-\sqrt{26}+1\end{cases}}\)

2, a, x^2-2x+1+4=(x-1)^2+4\(\ge\)4

b, 4x^2-4x+1-1+y^2+2y+1-1-2015=(2x-1)^2+(y+1)^2-2017\(\ge\)-2017

mk làm như thế thôi chứ bài kia dài quá mk làm biếng sory

12 tháng 7 2017

Nguyễn Thị Hà Tiên : Cảm ơn bạn nhiều lắm =)) Mik đã bt hướng làm bài rồi :3 Thực sự cảm ơn pạn nek <3