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

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

\(=-b\left(a-b\right)^2\)

a) 5a² - 5b² = 5(a²-b²) = ( a+b) (a-b)

b) 2ab² - a²b - b³ = b ( 2ab-a²-b² ) = -b (a²-2ab+b²) = -b (a-b)²

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

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

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

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

Xét tập hợp 1 :

x chia hết cho 2 và 3 => x chia hết cho 6 

A = { 6; 12; 18; ...; 60; 96 }

Vậy,......

a. (a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2

b. (a+b)^3= (a+b)(a+b)(a+b) = (a^2 + 2ab + b^2)(a + b) = a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 = a^3 + 3a^2b + 3b^2a + b^3

c. (a-b)^3= (a - b)(a-b)(a-b) = (a^2 - 2ab + b^2)(a - b) = a^3 - a^2b - 2a^2b + 2ab^2 + b^2a - b^3 = a^3 - 3a^2b + 3ab^2 - b^3

e. (a-b) ( a^2 + ab +b^2) = a^3 + a^2b + b^2a - ba^2 - ab^2 - b^3 = a^3 - b^3

g. ( a-b) ( a+b) = a^2 +ab -ab - b^2 = a^2 - b^2

tìm giá trị x để biểu thức nguyên

D=2x-3/x+5 

E=x^2-5/x-3

ai giup minh voi mai phai nop roi

câu 1 

xét tích 3 số

=(3a^2.b.c^3).(-2a^3b^5c).(-3a^5.b^2.c^2)

=[3.(-2).(-3)].(a^2.a^3.a^5).(b.b^5.b^2).(c.c^3.c^2)

=18.a^10.b^8.c^5 bé hơn hoặc bằng 0

=>tích 3 số đó không thể cùng âm=>3 số đó ko cùng âm dc

bây giờ mk đi học rùi tí về mk làm típ nhá

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

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

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

d: \(=x^2\left(2x+3\right)+2x+3=\left(2x+3\right)\left(x^2+1\right)\)

k mk đi

ai k mk

mk k lại

thanks

\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)

\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)

\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)

b.

Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)

\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)

\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)

b: \(=\left(x+1\right)^3-64y^3\)

\(=\left(x+1-4y\right)\left(x^2+2x+1+4xy+4y+16y^2\right)\)

c: \(=x\left(a^3-1\right)-b\left(a-1\right)\)

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

g: \(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)

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

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