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a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)
Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da
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a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
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b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
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a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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a: Ta có: \(71-\left(x+33\right)=26\)
\(\Leftrightarrow x+33=45\)
hay x=12
b: Ta có: \(\left(x+73\right)-26=76\)
\(\Leftrightarrow x+73=102\)
hay x=29
c: Ta có: \(45-\left(x+9\right)=6\)
\(\Leftrightarrow x+9=39\)
hay x=30
a) 71 - ( 33 + x ) = 26
(33 + x ) = 71 -26
33 + x = 45
x = 45 - 33 = 12
b) ( x + 73 ) - 26 = 76
( x + 73 ) = 102
x = 102 - 73 = 29
c) 45 - ( x + 9 ) = 6
( x + 9 ) = 45 - 6 = 39
x = 39 - 9 = 30
d) 89 - ( 73 - x ) = 20
73 - x = 89 - 20
73 - x = 69
x = 73 - 69 = 4
e) 4(x+41) = 400
x + 41 = 400 : 4 = 100
x = 100 - 41 = 59
f) 11(x-9 ) = 77
x - 9 = 77 : 11 = 7
x = 7 + 9 = 16
g) x + 7 = 25 + 13 = 38
x = 38 - 7 = 31
h) x + 4 = 198 - 120 = 78
x = 78 - 4 = 74
i) x - 9 = 350 : 5 = 70
x = 70 + 9 = 79
j) 2x - 49 = 5 . 9 = 45
2x = 45 + 49 = 94
x = 94 : 2 = 47
k) 25 + 3( x - 8 ) = 106
3(x-8 ) = 106 - 25 = 81
x - 8 = 81 : 3 = 27
x = 27 +8= 35
l) 9( x + 4 ) - 25 = 20
9( x + 4 ) = 20 + 25 = 45
x + 4 =45 : 9 = 5
x = 5 - 4 = 1
m) 200 - ( 2x + 6 ) = 64
2x + 6 = 200 - 64 = 136
2x = 136 - 6 = 130
x = 130 : 2 = 65
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\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
a, 2x+|x|=3x
|x|=3x-2x
|x|=x
=>x là số tự nhiên
a)2x+/x/=3x
/x/=3x-2x
/x/=1x
/x/=x
=>x=x
Vậy x=x là gtri cần tìm