-91/201 lớn hơn

Trả lời

\(\frac{-97}{201}\) > \(\frac{-194}{399}\)

HOK TOT~

Ta có \(\frac{12}{17}=1-\frac{5}{17}\)

\(\frac{13}{18}=1-\frac{5}{18}\)

Nhận thấy 17 < 18 

=> \(\frac{5}{17}>\frac{5}{18}\Leftrightarrow-\frac{5}{17}< -\frac{5}{18}\Rightarrow1-\frac{5}{17}< 1-\frac{5}{18}\)

=> \(\frac{12}{17}< \frac{13}{18}\)

\(\frac{12}{17}< \frac{13}{18}\)

15623 nha bn nhớ k mình nhé

\(5^6-2=15623\)

Đáp án là b)

10 người trả lời đúng trước mình k cho

Mk đoán là 0

a)5^x+1=125

=>5^x+1=5³

=>x+1=3

=>x=2

vậy x=2

b)4^2x+1=64

=>4^2x+1=4³

=>2x+1=3

=>x=1

vậy x =1

e)=>4^3x+2017=4^2020-3

=>3x+2017=2017

=>x=0

vậy x=0

f)=>2^x+2^x x 2³=144

=>2^x x (1+2³)=144

=>2^x  x   9=144

=>2^x=16

=>2^x=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 \(x=8\)

có bạn nào biết thì trả lời giùm mình nhá hi hi!

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 \(x=\frac{1}{4}\)

0.5 x = x - 8   

0.5 x - x = 8

-0.5 x = 8

x = 8 : ( -0.5 )

x = -16

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, \(B=\frac{1}{120}-\frac{2}{30.33}-\frac{2}{33.36}-...-\frac{2}{117.120}\)

\(\Rightarrow B=\frac{1}{120}-\left(\frac{2}{30.33}+\frac{2}{33.36}+...+\frac{2}{117.120}\right)\)

\(\Rightarrow B=\frac{1}{120}-\left(\frac{2.3}{30.33.3}+\frac{2.3}{33.36.3}+...+\frac{2.3}{117.120.3}\right)\)

\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{3}{30.33}+\frac{3}{33.36}+...+\frac{3}{117.120}\right)\)

\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{1}{30}-\frac{1}{33}+\frac{1}{33}-\frac{1}{36}+...+\frac{1}{117}-\frac{1}{120}\right)\)

\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\left(\frac{1}{30}-\frac{1}{120}\right)\)

\(\Rightarrow B=\frac{1}{120}-\frac{2}{3}\cdot\frac{1}{40}=\frac{1}{120}-\frac{1}{60}=-\frac{1}{120}\)