\(\frac{4}{3.6}+\frac{4}{6.9}+.........\frac{4}{27.30}\)
\(\frac{4}{3.6}\)+ \(\frac{4}{6.9}\)+ \(\frac{4}{9.12}\)+\(\frac{4}{12.15}\)
= 4(\(\frac{1}{3.6}\)+\(\frac{1}{6.9}\)+\(\frac{1}{9.12}\)+\(\frac{1}{12.15}\))
=\(\frac{4}{3}\)( \(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\) )
=\(\frac{4}{3}\)(\(\frac{1}{3}-\frac{1}{15}\))
=\(\frac{4}{3}\).\(\frac{4}{15}\)
=\(\frac{16}{45}\)
mk làm đúng chưa
tính
a) A=\(\frac{4^{16}.4^4}{2^{20}}\)
b) B=\(\frac{2^6.9^2}{6^{^4}.8}\)
c) C=\(\frac{6^3+3.6^2+3^3}{13}\)
a, \(A=\frac{4^{20}}{2^{20}}=\left(\frac{4}{2}\right)^{20}=2^{20}\)
\(A=\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+..........+\frac{5}{96.99}\)
3/5 A = 3/3.6 + 3/6.9 +..... + 3/96.99
= 1/3 - 1/6 + 1/6 - 1/9 + .... + 1/96 - 1/99 = 1/3 - 1/99 = 32/99
=> A = 160/297
k mk nha
\(\left(\frac{-1}{3.6}-\frac{1}{6.9}-\frac{1}{9.12}-\frac{1}{12.15}:\left|x\right|=\frac{-8}{15}\right)\)
\(F=\frac{2}{3.6}+\frac{2}{6.9}+\frac{2}{9.12}+...+\frac{2}{30.33}+\frac{3}{33.36}\)=?
tinh gia tri cua cac bieu thuc sau
\(\frac{20^{5.}5^{10}}{100^5}\)
b)\(\frac{6^3+3.6^2+3^3}{-13}\)
c)\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)l
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)
số số hạng của dãy\(\frac{1}{3.6};\frac{1}{6.9};\frac{1}{9.12};...;\frac{1}{156.159}\)là
Tính:
\(\frac{2}{3.6}+\frac{2}{6.9}+\frac{2}{9.12}+...+\frac{2}{117.120}\)
LƯU Ý: Dấu "." là nhân nhé !
Giúp mk với nha ,plz!!! Mk tick cho !!!! <3
Tính giá trị biểu thức
E=\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
F=\(\frac{6^3+3.6^2+3^3}{-13}\)
E = \(\frac{\left(2^2\right)^6.\left(3^2\right) ^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
E = \(\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
E = \(\frac{2^{12}.3^{10}+2^{13}.3^{10}.5}{-2^{11}.3^{11}.\left(2.3+1\right)}\)
E = \(\frac{2^{12}.3^{10}.\left(1+5\right)}{-2^{11}.3^{11}.7}\)
E = \(\frac{2^{12}.3^{10}.6}{-2^{11}.3^{11}.7}\)
E=\(\frac{-2^{11}.\left(-2\right).3^{10}.6}{-2^{11}.3^{10}.3.7}\)
E = \(\frac{-2.6}{3.7}=-\frac{4}{7}\)
Vậy E = -4/7
Ý F bn lm tương tự nha