\(\frac{101}{19}\)x \(\frac{61}{218}\)- \(\frac{101}{218}\)x \(\frac{42}{19}\)+ \(\frac{117}{218}\)
tính hợp lý
A=101/19*61/218-101/218*42/19+117/218
Tính hợp lý :
\(\left(5-\frac{43}{10}\right)-\left(\frac{42}{19}-\left(\frac{7}{2}-\frac{59}{10}\right)\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(\left(5-\frac{43}{10}\right)-\left(\frac{42}{19}-\left(\frac{7}{2}-\frac{59}{10}\right)\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(=5-\frac{43}{10}-\left(\frac{42}{19}-\frac{7}{2}+\frac{59}{10}\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(=5-\frac{43}{10}-\frac{42}{19}+\frac{7}{2}-\frac{59}{10}+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(=5-\frac{43}{10}+\frac{7}{2}+\frac{4}{5}\)
\(=5+\frac{35}{10}+\frac{8}{10}-\frac{43}{10}=5\)
Bài 1 :Thực hiện phép tính
a) N=1-5-9+13+17-21-25+......+2001-2005-2009+2013
b)So sánh P và Q
Biết P=\(\frac{2010}{2011}\)+\(\frac{2011}{2012}\)+\(\frac{2012}{2013}\)và Q=\(\frac{2010+2011+2012}{2011+2012+2013}\)
Bài 2:
TÍnh: N=\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^6}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
Bài 3
Cho a,b là các số nguyên thỏa mãn(\(^{a^2+b^2}\))chia hết cho 3.Chứng minh rằng a và b chia hết cho 3
Bài 1:
a. https://olm.vn/hoi-dap/detail/100987610050.html
b. Giống nhau hoàn toàn => P=Q
Chỉ biết thế thôi
a,1−5−9+13+17−21−25+29+......+2001−2005−2009+2013=(1−5)−(9−13)+(17−21)−(25−29)+.....+(2001−2005)−(2009−2013)=−4+4−4+4−......−4+4=0mik biết làm z thoy
Tính hợp lí :
\(B=\frac{20+\frac{19}{3}+\frac{19}{101}}{7+\frac{7}{13}+\frac{7}{19}+\frac{7}{101}}\)
Tinh cach bieu thuc sau
A=2+4-6-8+10+12-14-16+....+210+214-216-218
B=\(2^{2013}-2^{2012}-2^{2011}-2^{2010}-...-2-1\)
C=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
Số số hạng của A :
( 218 - 2 ) : 2 + 1 = 109
A = 2 + 4 - 6 - 8 + 10 + 12 - 14 - 16 + .... + 210 + 214 - 216 - 218
A = 2 - 6 + 4 - 8 + 10 - 14 + 12 - 16 + ... + 210 - 216 + 214 - 218
Đến đây bạn xem lại đề giùm mình , hiện tại đang ghép nhóm sao cho hiệu mỗi nhóm là -4 ( ghép đôi ) . Mà 109 ko chia hết cho 2
4 số cuối cùng 210 , 214 , 216 , 218 đã đủ cặp và ko hề dư ra số nào
????
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
a)Tính\(\frac{\left(17\frac{8}{19}-16\frac{9}{18}\right)\left(17,5+16\frac{17}{51}-32\frac{15}{22}\right)}{\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}}\)
b) Chứng tò rằng:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))
\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)
b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)
(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)
=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)
=1/101+1/102+...+1/200
https://olm.vn/hoi-dap/question/60726.html
