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(4,329 : 0,001 - 43,2 x 100) x 150 + 2018 + 2019 : 10000
= (4,329 x 1000 - 4320) x 150 + 2018 + 2019 : 10000
= (4329 - 4320) x 150 + 2018 + 2019 : 10000
= 9 x 150 + 2018 + 0,2019
= 1350 + 2018 + 0,2019
= 3368 + 0,2019
= 3368,2019
1.Xx3,7+Xx6,3=120
Xx(3,7+6,3)=120
Xx10=120
X=120:10
X=12
2.128xX-12xX-16xX=5208000
(128-12-16)xX=5208000
100xX=5208000
X=5208000:100
X=52080
3.5xX+3,75xX+125xX=20
(5+3,75+125)xX=20
128,8xX=20
X=128,8:20
X=6,44
Câu cuối cùng cậu ghi sai rồi! 3,75xX+125xX nhé
\(\dfrac{121212}{242424}+\dfrac{1313}{2626}+\dfrac{20212021}{40424042}\)
= \(\dfrac{12}{24}+\dfrac{13}{26}+\dfrac{2021}{4042}\)
= \(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)
= \(\dfrac{3}{2}\)
121212/242424+1313/2626+20212021/404244024
= 1/2+1/2+1/2
=3/2
2019 + 2019 : 0,5 + 2019 : 0,2 + 2019 : 0,125
= 2019 x 1 + 2019 x 2 + 2019 x 5 + 2019 x 8
= 2019 x ( 1 + 2 + 5 + 8 )
= 2019 x 16
= 32304
Trl :
2019 + 2019 : 0,5 + 2019 : 0,2 + 2019 : 0,125
= 2019 x 1 + 2019 x 2 + 2019 x 5 + 2019 x 8
= 2019 x ( 1 + 2 + 5 + 8 )
= 2019 x 16
= 32304
\(\dfrac{3}{4}\times X-1=\dfrac{1}{2}\times X\\ \dfrac{3}{4}\times X-\dfrac{1}{2}\times X=1\\ \left(\dfrac{3}{4}-\dfrac{1}{2}\right)\times X=1\\ \dfrac{1}{4}\times X=1\\ X=1:\dfrac{1}{4}\\ X=4\)
\(\dfrac{3}{4}\times x-1=\dfrac{1}{2}\times x\\ \dfrac{3}{4}\times x-\dfrac{1}{2}\times x=1\\ x\times\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=1\\ x\times\dfrac{1}{4}=1\\ x=1:\dfrac{1}{4}\\x=4\)
\(\dfrac{1}{4}+\dfrac{1}{3}\times X=\dfrac{1}{2}\\ \dfrac{1}{3}\times X=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\\ X=\dfrac{1}{4}:\dfrac{1}{3}=\dfrac{1}{4}\times3=\dfrac{3}{4}\)
14+13×�=1213×�=12−14=14�=14:13=14×3=3441+31×X=2131×X=21−41=41X=41:31=41×3=43
D=1/1.5+1/5.9+...+1/41.45
4D=4/1.5+4/5.9+...+4/41.45
4D=1-1/5+1/5-1/9+...+1/41-1/45
4D=1-1/45
D=44/45:4=11/45
Lời giải:
$\frac{2626}{1919}\times \frac{165165}{143143}\times \frac{383838}{606060}-\frac{1}{2019}$
$=\frac{26}{19}\times \frac{165}{143}\times \frac{38}{60}-\frac{1}{2019}$
$=\frac{26}{19}\times \frac{15}{13}\times \frac{19}{30}-\frac{1}{2019}$
$=1-\frac{1}{2019}=\frac{2018}{2019}$