tìm x: x+5=15+x
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{x+6}{15}=\dfrac{5-x}{5}\)
\(\Leftrightarrow5x+30=75-15x\)
\(\Leftrightarrow x=\dfrac{9}{4}\)
\(\dfrac{x+6}{15}=\dfrac{5-x}{5}\)
<=> \(\dfrac{x+6}{15}=\dfrac{3\left(5-x\right)}{15}\)
<=> x + 6 = 3(5 - x)
<=> x + 6 = 15 - 3x
<=> x + 3x = 15 - 6
<=> 4x = 9
<=> x = \(\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{15}{19}\times\dfrac{38}{5}< x< \dfrac{67}{15}-\dfrac{56}{16}\\ \dfrac{5\times3\times19\times2}{19\times5}< x< \dfrac{67\times16-56\times15}{15\times16}\\ 3\times2< x< \dfrac{232}{240}\\ 6< x< \dfrac{232}{240}\) (*)
Mà : \(\dfrac{232}{240}< \dfrac{240}{240}=1\\ \)
Nên (*) là vô lí ( Do `6>232/240` )
Vậy \(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{4}{7}\times x=\dfrac{5}{9}+\dfrac{1}{5}=\dfrac{34}{45}\)
\(x=\dfrac{34}{45}:\dfrac{4}{7}=\dfrac{119}{90}\)
b) \(\dfrac{11}{15}:x=\dfrac{2}{3}-\dfrac{9}{15}=\dfrac{1}{15}\)
\(x=\dfrac{11}{15}:\dfrac{1}{15}=11\)
a) 4/7 x X - 1/5 = 5/9
4/7 x X = 5/9 + 1/5
4/7 x X = 34/45
X = 34/45 : 4/7
X = 119/90
b)9/15 + 11/15 : X = 2/3
11/15 : X = 2/3 - 9/15
11/15 : X = 1/15
X = 11/15 : 1/5
X = 11
![](https://rs.olm.vn/images/avt/0.png?1311)
b: =>x(8-7)=-33
=>x=-33
c: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
d: =>-2x-2-x+5+2x=0
=>3-x=0
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
(X-15)5=(x-15)3
(x-15)5-(x-15)3=0
(X-15)5-3=0( bạn có thể bỏ bước này)
(X-15)2=0
=>x-15=0
=>x=15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3\left(x-5\right)-4\left(x-3\right)=-12\)
\(\Leftrightarrow3x-15-4x+12=-12\)
\(\Leftrightarrow-x=-9\)
\(\Leftrightarrow x=9\)
Vậy x=9
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài : \(-\dfrac{7}{15}+\dfrac{6}{15}< x>\dfrac{4}{5}+\dfrac{11}{5}\)
Ta có :
\(-\dfrac{7}{15}+\dfrac{6}{15}=\dfrac{-7+6}{15}=-\dfrac{1}{15}\)
\(\dfrac{4}{5}+\dfrac{11}{5}=\dfrac{4+11}{5}=\dfrac{15}{5}=3\)
\(\Rightarrow-\dfrac{1}{15}< x>3\)
Vì \(x\in N\) nên từ -1 đến \(-\dfrac{1}{15}\) loại
Vậy \(x>3\Rightarrow x\in\left\{3;.....\right\}\)