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29 tháng 6 2023

a) \(a-5\sqrt{a}\)

\(=\sqrt{a}\left(\sqrt{a}-\sqrt{5}\right)\)

b) \(a-7\)

\(=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)

c) \(a+4\sqrt{a}+4\)

\(=\left(\sqrt{a}+2\right)^2\)

d) \(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12\)

\(=\sqrt{x}\left(\sqrt{y}-4\right)+3\left(\sqrt{y}-4\right)\)

\(=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)

a: \(a-5\sqrt{a}=\sqrt{a}\left(\sqrt{a}-5\right)\)

b: \(a-7=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)

c: \(a+4\sqrt{a}+4=\left(\sqrt{a}+2\right)^2\)

d: \(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12\)

=căn x(căn y-4)+3(căn y-4)

=(căn y-4)(căn x+3)

8 tháng 7 2021

a)\(a-5\sqrt{a}=\sqrt{a}\left(\sqrt{a}-5\right)\)

b)\(a-7=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)

c)\(a+4\sqrt{a}+4=\left(\sqrt{a}+2\right)^2\)

d)\(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12=\sqrt{x}\left(\sqrt{y}-4\right)+3\left(\sqrt{y}-4\right)=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)

8 tháng 7 2021

em cảm ơn ạ yeu

a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)

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

22 tháng 8 2023

a) \(4x^2-1\)

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

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

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(9x^2-\dfrac{1}{4}\)

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

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

d) \(\left(x-y\right)^2-4\)

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

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

e) \(9-\left(x-y\right)^2\)

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

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

f) \(\left(x^2+4\right)^2-16x^2\)

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

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

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

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

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

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

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

3 tháng 12 2021

\(a,=6y\left(2x^2-3xy-5y^2\right)\\ =6y\left(2x^2+2xy-5xy-5y^2\right)\\ =6y\left(x+y\right)\left(2x-5y\right)\\ b,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ c,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\\ =\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ d,=\left(a^2+3b\right)^2-1=\left(a^2+3b+1\right)\left(a^2+3b-1\right)\\ e,=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\\ =\left(2x-5\right)\left(2x+5-2x-7\right)\\ =-2\left(2x-5\right)\\ f,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ g,=x^3-x-6x-6\\ =x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ =\left(x+1\right)\left(x^2-3x+2x-6\right)\\ =\left(x+1\right)\left(x-3\right)\left(x+2\right)\\ l,=x^4+4x^2+4-4x^2\\ =\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\\ h,=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)

12 tháng 10 2021

a) Sửa đề: \(a^2x+a^2y-7x-7y\)

\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)

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

 

12 tháng 10 2021

\(c,Sửa:x^2-2x+2y-y^2=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\\ d,=\left(4x^4+36x^2+81\right)-36x^2\\ =\left(2x^2+9\right)^2-36x^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\\ e,=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x^2+x-x+1\\ =x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

2 tháng 10 2021

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)