Tìm x, biết: x + 1/5 - 3/7 = 6/35
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a: =>x=3/7+3/5=15/35+21/35=36/35
b: =>x/35=4/5-5/7=28/35-25/35=3/35
=>x=3
c: =>x<3/4+8/4=11/4
=>\(x\in\left\{0;1;2;3\right\}\)
d: =>5/3<x<5/6+24/6=29/6
=>\(x\in\left\{2;3;4\right\}\)
e: =>x<10/12-9/12=1/12
=>x=0
f: =>2/3<x<12/6-5/6=7/6
=>x=1
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\(a,\frac{3}{35}-\left[\frac{3}{5}+x\right]=\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{2\cdot5}{35}\)
\(\Rightarrow\frac{3}{5}+x=\frac{3}{35}-\frac{10}{35}=\frac{-7}{35}\)
\(\Rightarrow x=\frac{-7}{35}-\frac{3}{5}=\frac{-7}{35}-\frac{3\cdot7}{35}=\frac{-7}{35}-\frac{21}{35}=\frac{-28}{35}=\frac{-4}{5}\)
\(b,\frac{5}{6}+\frac{1}{6}:x=\frac{3}{4}\)
\(\Rightarrow\frac{1}{6}:x=\frac{3}{4}-\frac{5}{6}\)
\(\Rightarrow\frac{1}{6}:x=\frac{18}{24}-\frac{20}{24}\)
\(\Rightarrow\frac{1}{6}:x=\frac{-2}{24}\)
\(\Rightarrow x=\frac{1}{6}:\frac{-2}{24}=\frac{1}{6}:\frac{-1}{12}=\frac{12}{-6}=\frac{-12}{6}=-2\)
\(c,\left|x-3\frac{4}{7}\right|=-0,5\)
Vì \(\left|x-3\frac{4}{7}\right|\ge0\)mà \(-0,5< 0\)nên \(x\in\varnothing\)
Vậy không có giá trị x thỏa mãn
a) 3/35-(3/5+x)=2/7
<=> 3/5+x=3/35-2/7
<=> 3/5+x=3/35-10/35
<=> 3/5+x=-7/35
<=> 3/5+x=-1/5
<=> x=-1/5-3/5
<=> x=-4/5
Vậy x=-4/5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{a}\cdot\frac{3}{7}\cdot\frac{1}{5}=\frac{6}{35}\)
\(\frac{2\cdot3\cdot1}{a\cdot7\cdot5}=\frac{6}{35}\)
\(\Rightarrow\frac{6}{a\cdot35}=\frac{6}{35}\)
\(\Rightarrow a\cdot35=35\)
\(a=35:35=1\)
P/s: Ủa rồi tìm x hay tìm a vậy bạn :D???
2/a x 3 / 7 x 1 / 5 = 6 / 35
2 / a x 3 / 7 = 6 / 35 : 1 / 5
2 / a x 3 / 7 = 6 / 7
2 / a = 6 / 7 : 3 / 7
2 / a = 2
= > a = 1
Học tốt
-NGL
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a. (5.x + 3) : 4 – 16 = 6
(5.x + 3) : 4 = 22
(5.x + 3) = 88
5.x = 85
x = 13
b. 7(23 – 3.x) : 2 – 35 = 35
7(23 – 3.x) : 2 = 70
7(23 – 3.x) = 35
7 . 3.x = 12
7.x = 4
x = 4/7
a.) (5.x + 3) : 4 – 16 = 6
=> (5.x + 3) : 4 = 22
=> 5.x + 3 = 88
=> 5x = 85
=> x = 17
b) 7(23 – 3.x) : 2 – 35 = 35
=> 7(23 – 3.x) : 2 = 70
=>7(23 – 3.x) = 140
=> 23 – 3.x = 20
=> 3x = 3
=> x = 1
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a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{17}{35}\) + 0,4
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{3}{35}\)
\(x\) = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)
\(x\) = - \(\dfrac{1}{5}\)
b, 0,2.(\(x\) - 3) +2,4 = 10
0,2.(\(x\) - 3) = 10 - 2,4
0,2.(\(x\) - 3) = 7,6
\(x\) - 3 = 7,6:0,2
\(x\) - 3 = 38
\(x\) = 38 + 3
\(x\) = 41
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{3}{7}\)
\(x=\dfrac{7}{9}\)
\(x:\dfrac{6}{8}=\dfrac{2}{5}\)
\(x=\dfrac{2}{5}\times\dfrac{6}{8}\)
\(x=\dfrac{6}{20}=\dfrac{3}{10}\)
\(x-\dfrac{2}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{2}{3}\)
\(x=\dfrac{9}{6}=\dfrac{3}{2}\)
\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)
\(x=\dfrac{1}{7}+\dfrac{4}{35}+\dfrac{2}{5}\)
\(x=\dfrac{23}{35}\)
.
`x + 1/5 - 3/7 = 6/35`
`x + 7/35 - 15/35 = 6/35`
`x - 8/35 = 6/35`
`x=6/35 + 8/35`
`x= 14/35`
`x=2/5`
Vậy `x= 2/5`
x+1/5-3/7=6/35
x+1/5=6/35+3/7
x+1/5=3/5
x=3/5-1/5
x=2/5