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TN
1
NN
2
30 tháng 6 2021
a) Ta có: 3x-17=x+3
\(\Leftrightarrow3x-x=3+17\)
\(\Leftrightarrow2x=20\)
hay x=10
b) Ta có: \( \left|x-3\right|-12=\left|-5\right|\)
\(\Leftrightarrow\left|x-3\right|=17\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=17\\x-3=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-14\end{matrix}\right.\)
Vậy: \(x\in\left\{20;-14\right\}\)
c) Ta có: \(25-\left(x-5\right)=-415-\left(15-415\right)\)
\(\Leftrightarrow25-x+5=-415-15+415\)
\(\Leftrightarrow30-x=-15\)
hay x=45
Vậy: x=45
30 tháng 6 2021
3x - 17 = x + 3
3x = x+3 +17
3x = x+20
3x-x=20
2x = 20
x= 20:2
x = 10
b ) |x-3| -12 = |-5|
|x-3| = 12 + 5
|x-3|=17
\(\left[{}\begin{matrix}x-3=17\\x-3=-17\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=17-3\\x=-17-3\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=14\\x=-20\end{matrix}\right.\)
vậy x= 14 hoặc x= -20
c) 25 -(x-5) = -415 -(15-145)
25-(x-5) = 15
x-5 = 25 -15
x-5 = 10
x = 10 - 5
x = 5
OY
4 tháng 6 2021
a,\(\dfrac{2}{3}\)-\(\dfrac{5}{7}\).\(\dfrac{14}{25}\)
=\(\dfrac{2}{3}\)-\(\dfrac{2}{5}\)
=\(\dfrac{4}{15}\)
b,\(\dfrac{-2}{5}\).\(\dfrac{5}{8}\)+\(\dfrac{5}{8}\).\(\dfrac{3}{5}\)
=\(\dfrac{5}{8}\).(\(\dfrac{-2}{5}\)+\(\dfrac{3}{5}\))
=\(\dfrac{5}{8}\).\(\dfrac{1}{5}\)
=\(\dfrac{1}{8}\)
4 tháng 6 2021
Giải:
a) \(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{4}{15}\)
b) \(\dfrac{-2}{5}.\dfrac{5}{8}+\dfrac{5}{8}.\dfrac{3}{5}\)
\(=\dfrac{5}{8}.\left(\dfrac{-2}{5}+\dfrac{3}{5}\right)\)
\(=\dfrac{5}{8}.\dfrac{1}{5}\)
\(=\dfrac{1}{8}\)
c) \(\left(\dfrac{1}{2}\right)^2-1\dfrac{1}{2}+0,5.\dfrac{12}{5}+5\%\)
\(=\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}+\dfrac{1}{20}\)
\(=\dfrac{-5}{4}+\dfrac{6}{5}+\dfrac{1}{20}\)
\(=0\)
Chúc bạn học tốt!
4 tháng 6 2021
`B=(2015+2016+2017)/(2016+2017+2018)`
`=2015/(2016+2017+2018)+2016/(2016+2017+2018)+2017/(2016+2017+2018)`
Vì `2015/(2016+2017+2018)<2015/2016`
`2016/(2016+2017+2018)<2016/2017`
`2017/(2016+2017+2018)<2017/2018`
`=>B<A`
OY
4 tháng 6 2021
Bài 5
B= \(\dfrac{2015}{2016+2017+2018}\)+\(\dfrac{2016}{2016+2017+2018}\)+\(\dfrac{2017}{2016+2017+2018}\)
Ta có:\(\dfrac{2015}{2016}\)>\(\dfrac{2015}{2016+2017+2018}\),\(\dfrac{2016}{2017}\)>\(\dfrac{2016}{2016+2017+2018}\),\(\dfrac{2017}{2018}\)>\(\dfrac{2017}{2016+2017+2018}\)
⇒A>B
NT
2
mình đg cần rất gấp mong ai giúp với
KM
1
ta có
\(\frac{1}{42}=\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7},\frac{1}{56}=\frac{1}{7}-\frac{1}{8};...;\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Nên ta có :\(\frac{1}{42}+\frac{1}{56}+..+\frac{1}{n\left(n+1\right)}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+..+\frac{1}{n}-\frac{1}{n+1}\)
\(=\frac{1}{6}-\frac{1}{n+1}=\frac{665}{4002}\Rightarrow\frac{1}{n+1}=\frac{1}{2001}\Rightarrow n=2000\)
b.ta có :\(A=\frac{1}{6^2}+\frac{1}{9^2}+..+\frac{1}{3033^2}=\frac{1}{9}\left(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{1011^2}\right)\)
\(< \frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{1010.1011}\right)=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1010}-\frac{1}{1011}\right)\)
\(=\frac{1}{9}\left(1-\frac{1}{1011}\right)< \frac{1}{9}\)
Vậy A<1/9