So sánh:
\(\frac{-97}{201}\)và\(\frac{-194}{399}\)
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Ta có \(\frac{12}{17}=1-\frac{5}{17}\)
\(\frac{13}{18}=1-\frac{5}{18}\)
Nhận thấy 17 < 18
=> \(\frac{5}{17}>\frac{5}{18}\Leftrightarrow-\frac{5}{17}< -\frac{5}{18}\Rightarrow1-\frac{5}{17}< 1-\frac{5}{18}\)
=> \(\frac{12}{17}< \frac{13}{18}\)
a)5x+1=125
=>5x+1=53
=>x+1=3
=>x=2
vậy x=2
b)42x+1=64
=>42x+1=43
=>2x+1=3
=>x=1
vậy x =1
e)=>43x+2017=42020-3
=>3x+2017=2017
=>x=0
vậy x=0
f)=>2x+2x x 23=144
=>2x x (1+23)=144
=>2x x 9=144
=>2x=16
=>2x=24
=>x=4
vậy x=4
\(\left(2x-1\right).5^4=3.5^5\)
\(2x-1=3.\left(5^5:5^4\right)\)
\(2x-1=3.5\)
\(2x-1=15\)
\(2x=15+1\)
\(2x=16\)
\(x=16:2\)
\(x=8\)
Vậy x = 8
\(a)\left(2x-1\right).5^4=3.5^5\)
\(\left(=\right)\left(2x-1\right).625=3.3125\)
\(\left(=\right)\left(2x-1\right).625=9375\)
\(\left(=\right)2x-1=9375:625\)
\(\left(=\right)2x-1=15\)
\(\left(=\right)2x=15+1\)
\(\left(=\right)2x=16\)
\(\left(=\right)x=16:2\)
\(\left(=\right)x=8\)
Vậy \(x=8\)
Vì là tia phân giác nên hai tia đó sẽ chia đều các góc và bắt đầu từ cùng 1 điểm và nằm đối nhau:
\(\Rightarrow\)hai tia phân giác của 2 góc đối đỉnh đối nhau.
\(x-\frac{1}{8}=50\%\cdot x\)
\(x-\frac{1}{8}=\frac{1}{2}x\)
\(x-\frac{1}{2}x=\frac{1}{8}\)
\(x\cdot\left(1-\frac{1}{2}\right)=\frac{1}{8}\)
\(\frac{1}{2}x=\frac{1}{8}\)
\(x=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
Ta có S = \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{2013.2015}\)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{2015}\right)=\frac{3}{2}.\frac{2014}{2015}=\frac{3021}{2015}\)
b) B = \(\frac{1}{120}-\frac{2}{30.33}-\frac{2}{33.36}-...-\frac{2}{117.120}\)
\(=\frac{1}{120}-\frac{2}{3}\left(\frac{3}{30.33}+\frac{3}{33.36}+..+\frac{3}{117.120}\right)\)
\(=\frac{1}{120}-\frac{2}{3}\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-\frac{1}{36}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(=\frac{1}{120}-\frac{2}{3}\left(\frac{1}{30}-\frac{1}{120}\right)=\frac{1}{120}-\frac{2}{3}.\frac{1}{40}=\frac{1}{120}-\frac{2}{120}=-\frac{1}{120}\)
Trả lời:
a, \(S=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{2013.2015}\)
\(\Rightarrow S=\frac{3.2}{1.3.2}+\frac{3.2}{3.5.2}+\frac{3.2}{5.7.2}+...+\frac{3.2}{2013.2015.2}\)
\(\Rightarrow S=\frac{3}{2}\cdot\frac{2}{1.3}+\frac{3}{2}\cdot\frac{2}{3.5}+\frac{3}{2}\cdot\frac{2}{5.7}+...+\frac{3}{2}\cdot\frac{2}{2013.2015}\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)
\(\Rightarrow S=\frac{3}{2}\cdot\frac{2014}{2015}=\frac{3021}{2015}\)
b, \(B=\frac{1}{120}-\frac{2}{30.33}-\frac{2}{33.36}-...-\frac{2}{117.120}\)
\(\Rightarrow B=\frac{1}{120}-\left(\frac{2}{30.33}+\frac{2}{33.36}+...+\frac{2}{117.120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\left(\frac{2.3}{30.33.3}+\frac{2.3}{33.36.3}+...+\frac{2.3}{117.120.3}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{3}{30.33}+\frac{3}{33.36}+...+\frac{3}{117.120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-\frac{1}{36}+...+\frac{1}{117}-\frac{1}{120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{1}{30}-\frac{1}{120}\right)\)
\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\frac{1}{40}=\frac{1}{120}-\frac{1}{60}=-\frac{1}{120}\)
-91/201 lớn hơn
Trả lời
\(\frac{-97}{201}\) > \(\frac{-194}{399}\)
HOK TOT~