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\(a,\dfrac{2}{3}.x=\dfrac{2}{7}\\ x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{3}{7}\\ ---\\ b,x.\dfrac{3}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2}{3}\\ ---\\ c,x:\dfrac{8}{13}=\dfrac{13}{7}\\x=\dfrac{13}{7}.\dfrac{8}{13}=\dfrac{8}{7}\\ ----\\ d,\dfrac{3}{2}:x=\dfrac{7}{4}\\ x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{3}{2}.\dfrac{4}{7}=\dfrac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A)9-25=(7-x)-(25-7)
9-25=7-x-25-7
9-25=(7-7)-(x+25)
9-25=0-x-25
9-25=x-25
25+25=9+x
50=9+x
x=50-9
x=41
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3/8 . x = 9/8 - 1
3/8 . x = 1/8
x = 1/8 : 3/8
x = 1/3
b) 4/5 . x = 7/5 - 1/5
4/5 . x = 6/5
x = 6/5 : 4/5
x = 3/2
c) 12/7 : x + 2/3 = 7/5
12/7 : x = 7/5 - 2/3
12/7 : x = 11/15
x = 12/7 : 11/15
x = 180/77
d) 3.(x + 7) - 15 = 27
3.(x + 7) = 27 + 15
3.(x + 7) = 42
x + 7 = 42 : 3
x + 7 = 14
x = 14 - 7
x = 7
a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)
b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)
c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)
d) \(3\left(x+7\right)-15=27\)
\(\Leftrightarrow3\left(x+7\right)=42\)
\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 5 . 4x = 80
4x = 80 : 5
4x = 16
4x = 42
Vậy x = 2
b. 15 - |x| = 25
|x| = 15 - 25
|x| = -10
=> x rỗng vì giá trị tuyệt đối của một số phải là số nguyên dương
c. x2 - [62 - (82 - 9 . 7)3 - 7 . 5]3 - 5 . 3 = 13
x2 - [36 - (64 - 63)3 - 35]3 - 15 = 1
x2 - [36 - 13 - 35]3 - 15 = 1
x2 - [36 - 1 - 35]3 - 15 = 1
x2 - 03 - 15 = 1
x2 - 0 - 15 = 1
x2 - 0 = 1 + 15
x2 - 0 = 16
x2 = 16 + 0
x2 = 16
x2 = 42
Vậy x = 4
d. (x - 7)3 = 25 . 52 + 2 . 102
(x - 7)3 = 32 . 25 + 2 . 100
(x - 7)3 = 800 + 200
(x - 7)3 = 1000
(x - 7)3 = 103
x - 7 = 10
x = 10 + 7
x = 17
Vậy x = 17
b 15-|x|=25
|x|=15-25
|x|=-10
Suy ra x=-10 hoặc x=10
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2.\left(x+3\right)-3x=-2x+7\)
\(\Leftrightarrow2x+6-3x=-2x-7\)
\(\Leftrightarrow-x+6=-2x+7\)
\(\Leftrightarrow x=-6+7\)
\(\Leftrightarrow x=1\)
\(b,5.\left(x-3\right)-29=-2.\left(7-x\right)-23\)
\(\Leftrightarrow5x-15-29+23=14-2x\)
\(\Leftrightarrow5x-21=14-2x\)
\(\Leftrightarrow5x+2x=35\)
\(\Leftrightarrow x=5\)
\(c,-17+\left|5-x\right|=-2.7\)
\(\Leftrightarrow-17+\left|5-x\right|=-14\)
\(\Leftrightarrow\left|5-x\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=8\end{cases}}\)
\(d,21-\left|x+7\right|=-42:\left(-2\right)\)
\(\Leftrightarrow21-\left|x+7\right|=21\)
\(\Leftrightarrow\left|x+7\right|=0\)
\(\Leftrightarrow x+7=0\)
\(\Leftrightarrow x=7\)
\(f,\left|3x-5\right|-\left(-15\right)=5\)
\(\Leftrightarrow\left|3x-5\right|+15=5\)
\(\Leftrightarrow\left|3x-5\right|=-10\)
Vì \(\left|a\right|\ge0\)mà \(-10< 0\)nên \(x\in\Phi\)
Đã giải quyết xong!!!
giúp mình với: so sánh -37/56 với -377/567 ai giải đc mình cho
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,\left|x-5\right|=\left|-7\right|\\ \left|x-5\right|=7\\ \Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\\ Vậy...\\ c,\left|x-3\right|=\left|5\right|+\left|-7\right|\\ \left|x-3\right|=5+7\\ \left|x-3\right|=12\\ \Rightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\\Vậy...\\ d,\left(7-x\right)-\left(25+7\right)=-25\\ 7-x-32=-25\\ 7-x=-25+32\\ 7-x=7\\ x=7-7\\ x=0\\ Vậy...\\ e,\left|x-3\right|=7-\left(-2\right)\\ \left|x-3\right|=9\\ \Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\\ Vậy...\)