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\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a) 2x + 5 = 3⁴ : 3²
2x + 5 = 3²
2x + 5 = 9
2x = 9 - 5
2x = 4
x = 4 : 2
x = 2
b) (3x - 24).73 = 2.74
(3x - 24).73 = 148
3x - 24 = 148/73
3x = 148/73 + 24
3x = 1900/73
x = 1900/73 : 3
x = 1900/219
c) [3.(42 - x)] + 15 = 23.3
126 - 3x + 15 = 69
141 - 3x = 69
3x = 141 - 69
3x = 72
x = 72 : 3
x = 24
d) 126 + (132 - x) = 300
132 - x = 300 - 126
132 - x = 174
x = 132 - 174
x = -42
2)
a) 120 - (x + 55) = 60
x + 55 = 120 - 60
x + 155 = 60
x = 60 - 55
x = 5
b) (7x - 11).3 = 25.52 + 200
(7x - 11).3 = 1500
7x - 11 = 1500 : 3
7x - 11 = 500
7x = 500 + 11
7x = 511
x = 511 : 7
x = 73
c) 2x + 2x + 4 = 544
4x = 544 - 4
4x = 540
x = 540 : 4
x = 135
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1:
=>2x-3=0 hoặc 5/2-x=0
=>x=3/2 hoặc x=5/2
2: =>x=1/2+12=12,5
3: =>(2x+3/5-3/5)(2x+3/5+3/5)=0
=>2x(2x+6/5)=0
=>x=0 hoặc x=-3/5
4: =>-1/6x=-1/3
=>x=1/3:1/6=2
5: =>1/4:x=1/4
=>x=1
6: =>2/5x+11/15=1
=>2/5x=4/15
=>x=2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-5+x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5+x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{5;7\right\}\)
2) Ta có: \(\left(30-x\right)\left(2x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}30-x=0\\2x-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-30\\2x=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=8\end{matrix}\right.\)
Vậy: \(x\in\left\{30;8\right\}\)
3) Ta có: \(\left(-5-x\right)\left(17+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5-x=0\\17+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=5\\x=0-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-17\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;-17\right\}\)
4) Ta có: \(\left(-3x+18\right)\left(-5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+18=0\\-5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-18\\-5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-2\right\}\)
Bài nay ta có hai vế bạn hãy đặt giả sử một trong hai vế bằng 0 rồi giải phương trình cho mỗi vế bằng o
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2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)
3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)
4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)
5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)
6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Rightarrow x+35=103\\ \Rightarrow x=68\\ b,\Rightarrow x+7=42\\ \Rightarrow x=35\\ c,\Rightarrow x-15=9\\ \Rightarrow x=24\\ d,\Rightarrow x+23=31\\ \Rightarrow x=8\\ e,\Rightarrow18-x=6\\ \Rightarrow x=12\\ f,\Rightarrow x+14=34\\ \Rightarrow x=20\\ g,\Rightarrow2x+1=21\\ \Rightarrow x=11\\ h,\Rightarrow3\left(x+1\right)=99\\ \Rightarrow x+1=33\Rightarrow x=32\\ i,5\left(x-3\right)=25\\ \Rightarrow x-3=5\\ \Rightarrow x=8\\ j,\Rightarrow8\left(2x+7\right)=88\\ \Rightarrow2x+7=11\Rightarrow x=2\\ k,\Rightarrow5\left(x+4\right)=85\\ \Rightarrow x+4=17\\ \Rightarrow x=13\)
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a, ( x + 1 ) + ( x + 2 ) + ... + ( x + 199 ) = 0
x + 1 + x + 2 + ... + x + 199 = 0
( x + x + ... + x ) + ( 1 + 2 + ... + 199 ) = 0
199x + 19900 = 0
199x = 0 - 19900
199x = -19900
x = -19900 : 199
x = -100
Vậy ...
b, ( x - 30 ) + ( x - 29 ) + ( x - 28 ) = 11
x - 30 + x - 29 + x - 28 = 11
( x + x + x ) - ( 30 + 29 + 28 ) = 11
3x - 87 = 11
3x = 11 + 87
3x = 98
x = \(\frac{98}{3}\)
Vậy ...
x =0 vì 011 =0
2x + 8 =20
2x =20-8=12
x = 6
5x -25 -2x =23
3x = 23+25 = 48
x = 48/3 = 16
câu 1 x=0
câu 2 x=7
câu 3 x= 1
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