\(B=-\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)

\(=-\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)

\(D=\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+3^9.2^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\\ =\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-6^{12}-6^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{6^{11}\left(-6-1\right)}=\dfrac{2^{12}.3^{10}.6}{6^{11}.-7}=\dfrac{2^{12}.3^{10}.2.3}{6^{11}.-7}\\ =\dfrac{2^{13}.3^{11}}{6^{11}.-7}=\dfrac{2^{11}.3^{11}.2^2}{6^{11}.-7}=\dfrac{6^{11}.4}{6^{11}.-7}=-\dfrac{4}{7}\)

D=4^6.9^5+6^9.120/-8^4.3^12-6^11
= 2^12.3^10+6^10.20/(-2)^12.3^12-6^11
= 20/3^2.6=10/27

a) 3125

b) -27

c) \(\frac{46}{5}\) hay 9,2

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)

=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)=     0

Đề sai nhé bạn

Số cuối chưa rõ ràng 

\(\frac{7}{11}:\frac{3}{5}+\frac{4}{11}x\frac{5}{6}\)

=\(\frac{7}{11}x\frac{5}{3}+\frac{20}{66}\)

=\(\frac{35}{33}+\frac{10}{33}\)

=\(\frac{45}{33}\)

=\(\frac{15}{11}\)

= 7/11 x 5/3 + 4/11 x 5/6

= 35/33 + 10/33

=15/11

CHÚC BẠN HỌC TỐT