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24 tháng 2 2019
cái thứ nhất mẫu số chung là 1898
cái thứ hai mẫu số chung là 360
17 tháng 3 2018
\(\dfrac{-14}{21};\dfrac{-2}{15};\dfrac{14}{-35}\)
\(\dfrac{-17}{21}=\dfrac{-85}{105}\);\(\dfrac{-2}{15}=\dfrac{-14}{105};\dfrac{14}{-35}=\dfrac{-14}{35}=\dfrac{-42}{105}\)
\(\dfrac{17}{60};\dfrac{5}{12};\dfrac{64}{90}\)
\(\dfrac{17}{60}=\dfrac{51}{180};\dfrac{-5}{12}=\dfrac{-75}{180};\dfrac{-64}{90}=\dfrac{-32}{45}=\dfrac{-128}{180}\)
bài2:
a)\(\dfrac{3}{5}>\dfrac{4}{7}\)
b)\(\dfrac{-5}{8}< \dfrac{-7}{12}\)
c)\(\dfrac{5}{-3}< \dfrac{-9}{12}\)
HD
1
XO
4 tháng 7 2019
a)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
= \(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
b) Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-...-\frac{1}{2^6}+\frac{1}{2^6}-\frac{1}{2^7}\)
A =\(1-\frac{1}{2^7}\)
4 tháng 7 2019
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=1-\frac{1}{11}\)
\(A=\frac{10}{11}\)
Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\left(1\right)\)
\(2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\frac{2}{2^5}+\frac{2}{2^6}+\frac{2}{2^7}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\left(2\right)\)
Lấy \(\left(2\right)-\left(1\right)\)hay \(2B-B\)ta có:
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(\Rightarrow B=1-\frac{1}{2^7}\)
\(\Rightarrow B=\frac{2^7-1}{2^7}=\frac{128-1}{128}=\frac{127}{128}\)
HOK TOT
PL
1 tháng 8 2018
Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)
\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)
\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)
1 tháng 8 2018
b) 3x - 7.(5x-1) = 6 - 2.(4-3x)
=> 3x - 35x + 7 = 6 - 8 + 6x
=> 3x - 35x - 6x = 6-8 -7
-38x = -9
x = 9/38
NN
2
CB
22 tháng 1 2018
a)\(\frac{24}{146}=\frac{24.13}{146.13}=\frac{312}{1898}\)
\(\frac{6}{13}=\frac{6.146}{13.146}=\frac{876}{1898}\)
b) \(\frac{7}{30}=\frac{7.60.40}{30.60.40}=\frac{16800}{72000}\)
\(\frac{13}{60}=\frac{13.30.40}{60.30.40}=\frac{15600}{72000}\)
\(\frac{-9}{40}=\frac{-9.30.60}{40.30.60}=\frac{-16200}{72000}\)
c)\(\frac{17}{60}=\frac{17.18.90}{60.18.90}=\frac{27540}{97200}\)
\(\frac{-5}{18}=\frac{5.60.90}{18.60.90}=\frac{27000}{97200}\)
\(\frac{-64}{90}=\frac{-64.18.60}{90.18.60}=\frac{-69120}{97200}\)
Mệt quá đi
5x+64:4=90
=>5x+16=90
=>5x=90-16=74
=>\(x=\frac{74}{5}\)
5\(x+64\) : 4 = 90
5\(x\) + 16 = 90
5\(x\) = 90 - 16
5\(x\) = 74
\(x\) = 74: 5
\(x=\frac{74}{5}\)
Vậy \(x=\frac{74}{5}\)