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BC
31 tháng 3 2019
a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
S
2 tháng 7 2019
a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)
\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)
\(=\frac{12}{15}\)
\(=\frac{4}{5}\)
c, \(\frac{3}{8}.3\frac{1}{3}\)
\(=\frac{3}{8}.\frac{10}{3}\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)
\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)
\(=\frac{-3}{5}+\frac{-60}{10}\)
\(=\frac{-3}{5}+\frac{-30}{5}\)
\(=\frac{-33}{5}\)
e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)
\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)
\(=\frac{2}{5}.10\)
\(=4\)
f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.-14\)
\(=-6\)
~Study well~
#KSJ
NC
25 tháng 8 2020
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
25 tháng 8 2020
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
23 tháng 12 2018
a,.\(\frac{1}{2}+\frac{2}{3}.\sqrt{9}\)
=\(\frac{1}{2}+\frac{2}{3}.3\)
=\(\frac{1}{2}+2\)
=\(\frac{5}{2}\)
b,\(x+\frac{2}{5}=1\)
\(x=1-\frac{2}{5}\)
X =\(\frac{3}{5}\)
23 tháng 12 2018
a, \(\frac{1}{2}+\frac{2}{3}\cdot\sqrt{9}\)
\(\Leftrightarrow\frac{1}{2}+\frac{2}{3}\cdot3\)
\(=\frac{1}{2}+2\)
\(=\frac{5}{2}\)
b, \(x+\frac{2}{5}=1\)
\(x=1+\left(\frac{-2}{5}\right)\)
\(x=\frac{3}{5}\)
c, Ta có : \(A=\left|x+\frac{1}{2}\right|\ge0\)
\(\Rightarrow A_{min}=0\)
Dấu "=" xảy ra khi : \(\left|x+\frac{1}{2}\right|=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(A_{min}=0\Leftrightarrow x=-\frac{1}{2}\)
d, \(2x-5=0\Leftrightarrow2x=0+5\Leftrightarrow2x=5\Leftrightarrow x=\frac{5}{2}\)
Vậy tập hợp các nghiệm của phương trình \(2x-5=0\) là \(\left\{\frac{5}{2}\right\}\).
1 tháng 4 2019
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
TN
15 tháng 6 2019
a/ \(\left(\frac{-2}{3}\right)^4:24=\frac{16}{81}:24=\frac{2}{243}\)
b/ \(\left(\frac{3}{4}\right)^3.4^4=\frac{27}{64}.256=108\)
c/ \(\frac{3.0,8^5}{2,4^4}=\frac{3.0,32768}{33,1776}=\frac{0,98304}{33,1776}=\frac{4}{135}\)
d/ \(\frac{3^3-0,9^5}{2,7^4}=\frac{27-0,59049}{53,1441}=\frac{26,40951}{53,1441}=0,4969415231\)
e/\(\left(\frac{-7}{2}\right)^2+\left(\frac{-3}{4}\right)^3.64-\left(\frac{-61}{5}\right)\)
\(=\frac{49}{4}+\frac{-27}{64}.64+\frac{61}{5}\)
\(=12,25-27+12,2\)
\(=-2,55\)
f/ \(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.5^3.3^3}{2^3.3^3.5^2.2^2}\)
\(=1-\frac{2^5.5^3.3^3}{2^5.3^3.5^2}=1-\frac{5}{1}=-4\)
\(\)
chúc bạn học tốt
KK
22 tháng 3 2019
Đặt \(\frac{x}{3}=\frac{y}{5}=k\)=> \(x=3k\) ; \(y=5k\)
Khi đó, ta có: C = \(\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)
= \(\frac{5.3^2.k^2+3.5^2.k^2}{10.3^2.k^2-3.5^2.k^2}\)
= \(\frac{k^2.\left(5.9+3.25\right)}{k^2.\left(10.9-3.25\right)}\)
= 8
25 tháng 3 2019
\(\frac{x}{3}=\frac{y}{5}\Rightarrow5x=3y\)
\(C=\frac{3xy+5xy}{6xy-5xy}=\frac{8xy}{1xy}=8\)
cách này nhanh hơn không :v
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{22}\)
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\)=\(\frac{8}{22}=\frac{4}{11}\)
Hok tốt