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(10 - 4.x) + 120 : 8 = 5 mũ 2
(10-4.x) +120 : 8 = 25
(10-4.x)+120 = 25.8
(10-4.x)=200
4.x=200+10
x=210:4
x=52,5
\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)
\(\Leftrightarrow1+x\cdot3=-4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
Ta có:
10 - 4x + 120:23 = 1 + 42
=> 10 - 4x + 15 = 1 + 16
=> 4x - 5 = 17
=> 4x = 17 + 5 = 22
=> x = 22/4
Vậy: x = 22/4
\(a) \frac{4}{5}.x=\frac{8}{35}\)
\(\implies x= \frac{8}{35}:\frac{4}{5}\)
\(\implies x=\frac{8}{35}.\frac{5}{4}\)
\(\implies x=\frac{2}{7}\). 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 \(x=-\frac{5}{3}\)
(6x - 4x) + 120:23 = 14 + 42
2x + 15 = 17
2x = 17 - 15
2x = 2
x = 1
\(1+2+3+4+...+x=120\)
\(\Rightarrow\frac{\left(x+1\right).x}{2}=120\)
\(\Rightarrow\left(x+1\right).x=120.2\)
\(\Rightarrow\left(x+1\right).x=240\)
\(\Rightarrow\left(x+1\right).x=16.15\)
\(\Rightarrow x=15\)