NM
2
K
Khách
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KA
29 tháng 10 2021
A = ( 6888 : 56 - 112 ) . 152 _ 13 . 72 + 13 . 28
A = ( 123 - 121 ) . 152 - 13 . ( 72 + 28 )
A = 2 . 152 - 13 . 100
A = 13.100 - 304
A = 1300 - 304
A = 996
L
0
ML
2
KK
14 tháng 3 2019
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
KN
14 tháng 3 2019
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B
5 tháng 9 2019
a) \(9.x-2.x=\frac{6^{27}}{6^{25}}+\frac{48}{12}\)
\(\Leftrightarrow7x=6^2+4\)
\(\Leftrightarrow7x=36+4=40\)
\(\Leftrightarrow x=\frac{40}{7}\)
Vậy : \(x=\frac{40}{7}\)
b) \(11^x=5.x+\frac{5^{31}}{5^{29}}+3.2^2-10^0\)
\(\Leftrightarrow11^x=5x+5^2+12-1\)
\(\Leftrightarrow11^x=5x+36\)
\(\Rightarrow x\in\varnothing\)
NH
22 tháng 8 2020
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)
22 tháng 8 2020
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
BT
10 tháng 9 2017
A= 13;21;34
B= 37;70;135
C= 64;128;256
D= 22;29;37
E= 53;68;75
F= 127;255;511
G= 49;64;81
H= 324;841;2209
I= chịu
k cho mk nha!
10 tháng 9 2017
a, A={x thuộc các số nguyên tố |2<hoặc bằng x<hoặc bằng 7}
oặc A={x thuộc R |(x^2-5*x+6)*(x^2-12*x+35)=0}
b,B={x thuộc Z | -3<hoặc bằng x<hoặc bằng 3}
c,C={5*x thuộc Z |-1<hoặc bằng x<hoặc bằng 3}
9 tháng 2 2017
a) \(24\left(16-5\right)-16\left(24-5\right)\)
\(=24.16-24.5-16.24+16.5\)
\(=\left(24.16-16.24\right)-24.5+16.5\)
\(=5\left(-24+16\right)\)
\(=5\left(-8\right)\)
\(=-40\)
b) \(29\left(9-13\right)-19\left(29-13\right)\)
\(=29.9-29.13-19.29+19.13\)
\(=\left(29.9-19.29\right)-\left(29.13-19.13\right)\)
\(=29\left(9-19\right)-13\left(29-19\right)\)
= \(29.\left(-10\right)-13.10\)
\(=-290-130\)
\(=-420\)
c) \(31.\left(-18\right)+31.\left(-81\right)-31\)
\(=31\left[-18+\left(-81\right)\right]-31\)
\(=31.\left(-99\right)-31\)
\(=31\left(-99-1\right)\)
\(=31.\left(-100\right)\)
\(=-3100\)
d) \(-12.47+\left(-12\right).52+\left(-12\right)\)
\(=-12\left(47+52+1\right)\)
\(=-12.100\)
\(=-1200\)
e) \(\left(-6-2\right)\left(-6+2\right)\)
= \(-8.\left(-4\right)\)
\(=32\)
\(f\)) \(\left(-5+9\right)\left(-4\right)\)
\(=4.\left(-4\right)\)
\(=-16\)
\(A=\left(6888:56-11^2\right).152+13.72+13.28\)
\(A=\left(6888:56-121\right).152+13.72+13.28\)
\(A=\left(123-121\right).152+13.72+13.28\)
\(A=2.152+13.72+13.28\)
\(A=13.\left(72+28\right)+152.2\)
\(A=1300+304\)
\(A=1604\)
\(B=\left[5082\right]\left(17^{29}:17^{27}-16^2\right)+13.12\left[\right]:31+9^2\)
\(B=\left[5082\left(17-16^2\right)15.6\right];31+81\)
\(B=310:31+81\)
\(B=10+81\)
\(B=91\)
A=1604
B=91