Tìm x:
1) 1/3+3/35<x/210<4/7+1/3
2) 5/3+4/3<x<16/5+18/10
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a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
\(1)\frac{12}{7}\times \frac{7}{4}+\frac{35}{11}:\frac{245}{121}\)
\(=3+\frac{35}{11}\times \frac{121}{245}\)
\(=3+\frac{11}{7}\)
\(= 3\frac{1}{7}(=\frac{22}{7})\)
-(1/3)-1/15-1/35-...-1/x.(x+2)=20/-21
=>1/3+1/15+1/35+...+1/x(x+2)=20/21
=>1/2 * (2/3 + 2/15 + 2/35 + ... + 2/x(x+2) )=20/21
=>\(\dfrac{2}{1\cdot3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{20}{21}:\dfrac{1}{2}\)
=>\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{40}{41}\)
=>\(1-\dfrac{1}{x+2}=\dfrac{40}{41}\)
=>\(\dfrac{1}{x+2}=1-\dfrac{40}{41}=\dfrac{1}{41}\)
=>x+2=41=>x=39
(1/3)-1/15-1/35-...-1/x.(x+2)=20/-21
=>1/3+1/15+1/35+...+1/x(x+2)=20/21
=>1/2 * (2/3 + 2/15 + 2/35 + ... + 2/x(x+2) )=20/21
=>21⋅3+23.5+25.7+...+2x(x+2)=2021:1221⋅3+23.5+25.7+...+2x(x+2)=2021:12
=>1−13+13−15+15−17+...+1x−1x+2=40411−13+13−15+15−17+...+1x−1x+2=4041
=>1−1x+2=4041
Bài 1:
a: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{2}\\x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
= \(\dfrac{44}{105}\) < \(\dfrac{x}{210}\) <\(\dfrac{158}{105}\) = \(\dfrac{88}{210}< \dfrac{x}{210}< \dfrac{316}{210}\)
=> x = 89 -> 315
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{24}{35}\)
\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)
\(\frac{3}{10}-\frac{3}{2x+4}=\frac{24}{35}\)
\(\frac{3}{2x+4}=\frac{-27}{70}\)
tự làm nốt
a: =>31-x=60
=>x=-29
b: =>(x-140):35=280-270=10
=>x-140=350
=>x=490
c: =>(1900-2x):35=48
=>1900-2x=1680
=>2x=220
=>x=110
d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)
=>2x-1=11
=>x=6
e: =>(x+2)^5=4^5
=>x+2=4
=>x=2
f: =>3x-4=0 hoặc x-1=0
=>x=4/3 hoặc x=1
g: =>(2x-1)^2=49
=>2x-1=7 hoặc 2x-1=-7
=>x=-3 hoặc x=4
h: =>x(x+1)/2=78
=>x(x+1)=156
=>x=12
b) 35 - 5 * x = 20
- 5 * x = 20 - 35
-5 * x = -15
x = \(\frac{-15}{-5}\)
x = \(\frac{15}{5}\)
x = 3
\(\dfrac{1}{3}+\dfrac{3}{35}< \dfrac{x}{210}< \dfrac{4}{7}+\dfrac{1}{3}\)
Ta có: \(\dfrac{1}{3}+\dfrac{3}{35}=\dfrac{35+9}{105}=\dfrac{44}{105}\)
và \(\dfrac{4}{7}+\dfrac{1}{3}=\dfrac{12+7}{21}=\dfrac{19}{21}\)
=> \(\dfrac{44}{105}=\dfrac{44.2}{105.2}=\dfrac{88}{210}\)
=> \(\dfrac{19}{21}=\dfrac{19.10}{21.10}=\dfrac{190}{210}\)
=> \(\dfrac{88}{201}< \dfrac{x}{210}< \dfrac{190}{210}\)
=> Vậy x ∈ {89; 90; 91; 92; ... ; 188; 189}
2: =>3<x<16/5+9/5=5
=>x=4
1: =>70/210+18/210<x/210<120/210+70/210
=>88<x<190
hay \(x\in\left\{89;90;...;189\right\}\)