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x - 2.32 : 3 = 12
=> x - 18 : 3 = 12
=> x - 6 = 12
=> x = 6 + 12 = 18
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a) \(x\left(x-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x-1=0\)
+) \(x=0\)
+) \(x-1=0\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
b) \(\left(x+2\right)\left(x-4\right)=0\)
\(\Rightarrow x+2=0\) hoặc \(x-4=0\)
+) \(x+2=0\Rightarrow x=-2\)
+) \(x-4=0\Rightarrow x=4\)
Vậy \(x\in\left\{-2;4\right\}\)
c) \(x^3.x^2=2^8:2^3\)
\(\Rightarrow x^5=2^5\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
d) \(3^{x-3}-3^2=2.3^2\)
\(\Rightarrow3^{x-3}=18+9\)
\(\Rightarrow3^{x-3}=27\)
\(\Rightarrow3^{x-3}=3^3\)
\(\Rightarrow x-3=3\)
\(\Rightarrow x=6\)
Vậy \(x=6\)
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8.9 - 3.(x + 2.9) + 6=0
72 - 3x +18.3 + 6 =0
(72+18.3 +6 ) -3x =0
(72+54+6) - 3x = 0
132 - 3x = 0
132 =3x
x = 132 :3 =44
nhớ túch đúng cho mình nha =)))
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(x-2)^20-(x-2)^18=0
(x-2)^18[(x-2)^2-1]=0
suy ra (x-2)^18=0;(x-2)^2-1=0
x-2=0 ;(x-2)^2=1 ;(x-2)^2=-1
x=2 ;x-2=1 ;x-2=-1
x=3 x=1
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33 . 32x - 1 = 81
33 . 32x - 1 = 34
32x - 1 = 34 : 33 = 3
=> 2x - 1 = 1
2x = 2
x = 2 : 2 = 1
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a) \(\left(\left|x\right|+3\right):5-3=12\Leftrightarrow\left|x\right|+3=45\Leftrightarrow\orbr{\begin{cases}x=42\\x=-42\end{cases}}\)
b) \(86:\left[2\left(2x-1\right)^2-7\right]+4^2=2\cdot3^2\Leftrightarrow2\left(2x-1\right)^2-7=43\Leftrightarrow\left(2x-1\right)^2=25\Leftrightarrow\orbr{\begin{cases}2x-1=-5\\2x-1=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
a) \(\left(\left|x\right|+3\right)\div5-3=12\)
\(\left(\left|x\right|+3\right)\div5=12+3\)
\(\left(\left|x\right|+3\right)\div5=15\)
\(\left|x\right|+3=15.5\)
\(\left|x\right|+3=75\)
\(\left|x\right|=75-3\)
\(\left|x\right|=72\)
\(\Rightarrow\orbr{\begin{cases}x=72\\x=-72\end{cases}}\)
Vậy \(x\in\left\{72;-72\right\}\)
b) \(86\div\left[2,\left(2x-1\right)^2-7\right]+4^2=2.3^2\)
\(86\div\left[2.\left(2x-1\right)^2-7\right]+16=18\)
\(86\div\left[2.\left(2x-1\right)^2-7\right]=18-16\)
\(86\div\left[2.\left(2x-1\right)^2-7\right]=2\)
\(2.\left(2x-1\right)^2-7=86\div2\)
\(2.\left(2x-1\right)^2-7=43\)
\(2.\left(2x-1\right)^2=43+7\)
\(2.\left(2x-1\right)^2=50\)
\(\left(2x-1\right)^2=50\div2\)
\(\left(2x-1\right)^2=25\)
\(\left(2x-1\right)^2=5^2\)
\(\Rightarrow2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6\div2\)
\(x=3\)