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\(a) \frac{4}{5}.x=\frac{8}{35}\)
\(\implies x= \frac{8}{35}:\frac{4}{5}\)
\(\implies x=\frac{8}{35}.\frac{5}{4}\)
\(\implies x=\frac{2}{7}\). Vậy \(x=\frac{2}{7}\)
\(b) \frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(\implies \frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(\implies \frac{3}{5}x=\frac{9}{14}\)
\(\implies x=\frac{9}{14}:\frac{3}{5}\)
\(\implies x=\frac{9}{14}.\frac{5}{3} \)
\(\implies x=\frac{15}{14}\). Vậy \(x=\frac{15}{14}\)
\(c) x-25\% x=0,5\)
\(\implies 75\% x=0,5\)
\(\implies \frac{3}{4}x=\frac{1}{2}\)
\(\implies x=\frac{1}{2}:\frac{3}{4}\)
\(\implies x=\frac{1}{2}.\frac{4}{3}\)
\(\implies x=\frac{2}{3}\). Vậy \(x=\frac{2}{3}\)
\(d) (\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
Có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\implies 1+x:\frac{1}{3}=-4\)
\(\implies x:\frac{1}{3}=-5\)
\(\implies x=-5.\frac{1}{3}\)
\(\implies x=\frac{-5}{3}\). Vậy \(x=\frac{-5}{3}\)
~ Hok tốt a~
\(\frac{4}{5}.x=\frac{8}{35}\) \(\frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(x=\frac{8}{35}:\frac{4}{5}\) \(\frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(x=\frac{8}{35}.\frac{5}{4}\) \(\frac{3}{5}x=\frac{9}{14}\)
\(x=\frac{2}{7}\) \(x=\frac{9}{14}:\frac{3}{5}\)
Vậy \(x=\frac{2}{7}\) \(x=\frac{9}{14}.\frac{5}{3}\)
\(x-25\%x=0.5\) \(x=\frac{15}{14}\)
\(x-\frac{1}{4}x=\frac{1}{2}\) Vậy \(x=\frac{15}{14}\)
\(x\left(1-\frac{1}{4}\right)=\frac{1}{2}\) \(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(x=\frac{1}{2}:\frac{3}{4}\) \(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\) \(x=\frac{2}{3}\) \(\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
Vậy \(x=\frac{2}{3}\) \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=\left(-4\right)-1\)
\(x:\frac{1}{3}=-5\)
\(x=\left(-5\right).\frac{1}{3}\)
\(x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}\)
\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{25\cdot26}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{26}\)
\(\Rightarrow A=\frac{12}{26}=\frac{6}{13}\)
(1/24.25 + 1/25.26 + ... + 1/29.30) . 120 + x : 1/3 = -4
(1/24 - 1/25 + 1/25 - 1/26 + ... + 1/29 - 1/30) . 120 + x . 3 = -4
(1/24 - 1/30) . 120 + x . 3 = -4
(5/120 - 4/120) + x . 3 = -4
=> 1/120 . 120 + x . 3 = -4
=> 1 + x . 3 = -4
=> x . 3 = -4 - 1
=> x . 3 = -5
=> x = -5/3
Vậy x = -5/3
a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\Rightarrow1+x:\frac{1}{3}=-4\)
\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)
\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)
b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)
\(\Rightarrow x=4\)
\((\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+y:\frac{1}{3}=-4\)
Ta có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+y:\frac{1}{3}=-4\)
\(\implies 1+y:\frac{1}{3}=-4\)
\(\implies y:\frac{1}{3}=-5\)
\(\implies y=-5.\frac{1}{3}\)
\(\implies y=\frac{-5}{3}\). Vậy \(y=\frac{-5}{3}\)
~ Học tốt a~
\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)
\(\Rightarrow\frac{1}{120}.120+y:\frac{1}{3}=-4\)
\(\Rightarrow1+y:\frac{1}{3}=-4\)
\(\Rightarrow y:\frac{1}{3}=-4-1\)
\(\Rightarrow y:\frac{1}{3}=-5\)
\(\Rightarrow y=-5.\frac{1}{3}\)
\(\Rightarrow y=\frac{-5}{3}\)
Vậy \(y=\frac{-5}{3}\)
_Chúc bạn học tốt_
Lời giải:
\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}=\frac{25-24}{24.25}+\frac{26-25}{25.26}+...+\frac{30-29}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}=\frac{1}{120}\)
Vậy:
\(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=-5\)
\(x=-15\)
\(\left(\dfrac{1}{24.25}+\dfrac{1}{25.26}+...+\dfrac{1}{29.30}\right).120+x:\dfrac{1}{3}=-4\)
\(\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\)
\(\left(\dfrac{1}{24}-\dfrac{1}{30}\right).120+x:\dfrac{1}{3}=-4\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
25.26 - 25.(5-1)2
= 25.26 - 25.42
= 25.26 - 25.16
= 25. (26-16)
=25.10
=250
Chúc bạn học tốt!!