tính
a)\(\frac{5}{6}\cdot\frac{3}{4}:2\)
b) \(\frac{2}{5}\cdot\frac{4}{7}:2\)
tìm x biết
a, \(\frac{1}{2}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=4^x\)
b, \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
c,\(\left|4x+3\right|-\left|x-1\right|=7\)
mong các bạn giúp !!!
b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)
\(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)
Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)
Vậy x= 4
TÌM x biết:
a) \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\frac{5}{12}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{62}=4^x\)
b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
c)\(\left|4x+3\right|-\left|x-1\right|=7\)
a.4^7
b.8^5
c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu
Thực hiện phép tính hợp lí nếu có thể:
a/ \(\frac{6}{7}+\frac{1}{7}\cdot\frac{2}{7}+\frac{1}{7}\cdot\frac{5}{7}\)
b/\(\frac{2}{3}\cdot\frac{5}{7}\cdot\frac{-3}{8}\cdot\frac{11}{5}\)
c/\(11\frac{4}{7}-\left(2\frac{3}{5}+5\frac{4}{7}\right)\)
d/\(\frac{3}{4}-\frac{3}{4}\cdot\left(\frac{2}{3}+1\right)\)
e/\(0.5\cdot1\frac{1}{3}\cdot75\%:\frac{2}{5}+\frac{3}{5}\)
f/\(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot\left(-2\right)^2\)
g/\(1\frac{3}{8}+\left(\frac{-5}{6}+\frac{7}{12}\right):\frac{2}{3}\)
h/\(1\frac{1}{4}\cdot\frac{-3}{2}+50\%\cdot98\)
i/\(\left(2,09:1,1+4,5\right)\cdot\frac{5}{8}+4,32\)
kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi
trời ơi bài dễ thế này tự làm đi còn hỏi
Đã biết thì đã không hỏi. Đồ kazuto kirigaya xấu tính!
1. Tính tổng
\(\frac{1}{2}\cdot\frac{1}{3}\cdot+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
* LÀM NỐT *
#Louis
1/2.1/3+1/3.1/4+1/4.1/5+...+1/8.1/9
=1/2.3=1/3.4+1/4.5+...+1/8.9\
=1/2-1/3+1/3-1/4=1/4-1/5+...+1/8.1/9
=1/2-1/9
=9/18-2/18
=7/18
HỌC TỐT NHA BẠN
So sánh M và N, biết
\(M=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\)và \(N=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\)
M=(1.3.5.7.....99)/(2.4.6.8.....100)
số số hạng của tử = (99-1)/2 +1 = 50 -> 1.3.5.7....99= (99+1)*50/2 =2500
số số hạng của mẫu = (100-2)/2+1 =50 -> 2.4.6.8....100= (100+2)*50/2 =2550
--> M= 2500/2550 =50/51
Làm tương tự với N ta có kq N=51/52 ->M/N= 2600/2601 -> M<N
Vào hướng dẫn viết công thức, hình vẽ ở cuối trang tạo câu hỏi và chọn video đầu ấy
Tính giá trị biểu thức
a,\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
b,\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
c,\(A=81\cdot\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right]:\frac{158158158}{711711711}\)
d,\(A=\frac{5\cdot\left(2^2.3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
\(a,\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(b,5\cdot\frac{7}{5}\) \(c,\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(d,4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(e,\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(g,2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
) Ta có: 1615⋅−514⋅5424⋅56211615⋅−514⋅5424⋅5621
=1615⋅−514⋅94⋅83=1615⋅−514⋅94⋅83
=4⋅−13⋅47⋅3=4⋅−13⋅47⋅3
=12⋅−421=−4821=−167=12⋅−421=−4821=−167
b) Ta có: 5⋅75=355=75⋅75=355=7
c) Ta có: 17⋅59+59⋅17+59⋅3717⋅59+59⋅17+59⋅37
=59(17+17+37)=59(17+17+37)
=59⋅57=2563=59⋅57=2563
d) Ta có: 4⋅11⋅34⋅91214⋅11⋅34⋅9121
=4⋅11⋅3⋅94⋅121=2711=4⋅11⋅3⋅94⋅121=2711
e) Ta có: 34⋅169−75:−212034⋅169−75:−2120
=43+43=83=43+43=83
g) Ta có: 213−13⋅[−32+(23+0,4⋅5)]213−13⋅[−32+(23+0,4⋅5)]
=73−13⋅[−32+23+2]=73−13⋅[−32+23+2]
=73−13⋅76=73−13⋅76
=73−718=4218−718=3518
Cho M=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot....\cdot\frac{99}{100}\)
N=\(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot......\cdot\frac{100}{101}\)
a, Tính M\(\times\)N
b, CM M<\(\frac{1}{10}\)
tính nhanh
A= \(\frac{19}{23}\cdot\frac{-4}{7}-\frac{4}{23}\cdot\frac{2}{7}\)
B= \(\frac{3}{5}+\frac{2}{5}\cdot\frac{-11}{3}+\frac{2}{3}\cdot\frac{-2}{5}+\frac{14}{15}\)
a) A = \(\frac{19}{23}.\frac{-4}{27}-\frac{4}{23}.\frac{2}{7}\)
= \(\frac{19}{7}.\frac{-4}{23}+\frac{-4}{23}.\frac{2}{7}\)
= \(\frac{-4}{23}.\left(\frac{19}{7}+\frac{2}{7}\right)\)
= \(\frac{-4}{23}.3\)
= \(\frac{-12}{23}\)
b) B = \(\frac{3}{5}+\frac{2}{5}.\frac{-11}{3}+\frac{2}{3}.\frac{-2}{5}+\frac{14}{15}\)
= \(\frac{9+14}{15}+\frac{2}{5}.\frac{-11}{3}+\frac{-2}{3}.\frac{2}{5}\)
= \(\frac{23}{15}+\frac{2}{5}\left(\frac{-11}{3}+\frac{-2}{3}\right)\)
= \(\frac{23}{15}+\frac{2}{5}.\frac{-13}{3}\)
= \(\frac{23}{15}+\frac{-26}{15}\)
= \(\frac{-3}{15}=\frac{-1}{5}\)
\(\frac{\frac{1}{3}-\frac{5}{2}}{\frac{3}{4}-\frac{1}{2}}\cdot\frac{\frac{5}{6}+\frac{7}{3}}{1-\frac{5}{6}}\cdot\frac{-\frac{2}{5}+1}{\frac{2}{5}-1}\)