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\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0
\(B=-\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=-\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
a, 205.510/1005
=205.55.55/1005
=1005.55/1005
=55
=3125
b, (0,9)5/(0,3)6
=(0,3.3)5/0,36
=0,55.35/0,36
=35/0,3
=810
c, 63+3.62+33/-13
=(2.3)3+3.(3.2)2+33/-13
=23.33+3.32.22+33/-13
=33.23+33.22+33/-13
=33(23+22+1)/-13
=27.13/-13
=-27
d, 46.95+69.120/84.312-611
=(22)6.(32)5+(2.3)9.3.23.5/(23)4.312-(2.3)11
=212.310+29.39.3.23.5/212.312-211.311
=212.310+212.310.5/211.311.2.3-211.311
=212.310.(1+5)/211.311(6-1)
=212.310.6/211.311.5
=2.6/3.5
=12/15
=4/5
a. \(\frac{20^5.5^{10}}{100^5}\)= \(\frac{20^5.5^{10}}{20^5.5^5}\)= \(5^5\)=\(3125\)
b. \(\frac{0,9^5}{0,3^6}\)= \(\frac{0,9^5}{0,3^5.0,3}\)= \(\left(\frac{0,9}{0,3}\right).\frac{1}{0,3}\)= \(243.\frac{1}{0,3}\)= \(810\)
c.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
\(D=\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+3^9.2^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\\ =\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-6^{12}-6^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{6^{11}\left(-6-1\right)}=\dfrac{2^{12}.3^{10}.6}{6^{11}.-7}=\dfrac{2^{12}.3^{10}.2.3}{6^{11}.-7}\\ =\dfrac{2^{13}.3^{11}}{6^{11}.-7}=\dfrac{2^{11}.3^{11}.2^2}{6^{11}.-7}=\dfrac{6^{11}.4}{6^{11}.-7}=-\dfrac{4}{7}\)
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) \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{5^{10}.9^{10}.5^{20}}{5^{15}.5^{15}.3^{15}}=\frac{5^{30}.3^{20}}{5^{30}.3^{15}}=3^5=243\)
b) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)
\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=-3^3=-27\)
c) \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\frac{2.6}{3.5}=\frac{4}{5}\)
d) \(\frac{3.11+42}{5^3}=\frac{33+42}{5^3}=\frac{75}{5^3}=\frac{5^2.3}{5^3}=\frac{3}{5}\)
\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)
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)