5-1.25x=125 (x thuộc Z)
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Có : \(\dfrac{x}{5}=\dfrac{125}{x}\)
\(\Leftrightarrow x^2=5.125\)
\(\Leftrightarrow x=\pm25\)
Có : \(\dfrac{x}{5}=\dfrac{125}{x}\)
\(\Leftrightarrow x^2=5.125\)
\(\Leftrightarrow x=\pm25\)
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
a , 5-1 . 25x = 125
\(\frac{1}{5}.25^x=125\)
\(25^x=125.5=625\)
\(25^x=25^2\)
x = 2
b , 5x - | 9 - 7x | = 3
Với x < \(\frac{9}{7}\)
9 - 7x < 0
| 9 - 7x | = - ( 9 - 7x ) = 7x - 9
5x - ( 7x - 9 ) = 3
5x - 7x + 9 = 3
-2x = -6
x = -6 : (-2 )
x = 3
Với x >= \(\frac{7}{9}\)
9 - 7x >= 0
| 9 - 7x | = 9 - 7x
5x - 9 - 7x = 3
-2x - 9 = 3
-2x = 3 + 9
-2x = 12
x = 12 : ( -2 ) = -6
\(x^3=-125\)
\(x^3=\left(-5\right)^3\)
\(\Rightarrow x=-5\)
a, \(\left(3x-7\right)-\left(-5-7x\right)=4x+9\)
\(\Leftrightarrow3x-7+5+7x=4x+9\)
\(\Leftrightarrow3x+7x-4x=9+7-5\)
\(\Leftrightarrow\left(3+7-6\right)x=11\Rightarrow4x=11\Rightarrow x=\frac{11}{4}\notin Z\)
Vậy : \(x\in\varnothing\)
b, \(\left(3x-7\right)-\left(-5-7x\right)=5x+8\)
\(\Leftrightarrow3x-7+5+7x=5x+8\)
\(\Leftrightarrow3x+7x-5x=8+7-5\)
\(\Leftrightarrow\left(3+7-5\right)x=10\Rightarrow5x=10\Rightarrow x=10\div5=2\in Z\)
Vậy : x = 2
c, \(\left(3x-2\right)^3=125\Leftrightarrow\left(3x-2\right)^3=5^3\)
\(\Rightarrow3x-2=5\Rightarrow3x=5+2\Rightarrow3x=7\)
\(\Rightarrow x=7\div3\Rightarrow x=\frac{7}{3}\notin Z\)
Vậy \(x\in\varnothing\)
a, (3x-7)-(-5-7x)=4x+9
3x-7+5+7x=4x+9
3x+7x-4x=9+7-5
6x=11
x=11/6(thuộc Z t/m)
Vậy...
b, (3x-7)-(-5-7x)=5x+8
3x-7+5+7x=5x+8
3x+7x-5x=8+7-5
5x=10
x=2(thuộc Z t/m)
vây...
c, (3x-2)^3=125
(3x-2)^3=5^3
3x-2=5
x=7/3(thuộc Z thỏa mãn)
Vậy...
a) \(5^{-1}.25^x=125\)
\(\Rightarrow5^{-1}.5^{2x}=5^3\)
\(\Rightarrow5^{2x-1}=5^3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(|x+1|+|x+2|+|x+3|=4x\)
Vì \(\hept{\begin{cases}|x+1|\ge0\forall x\\|x+2|\ge0\forall x\\|x+3|\ge0\forall x\end{cases}}\)
\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\hept{\begin{cases}x+1>0\\x+2>0\\x+3>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)
\(\Rightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow x=6\)
Vậy \(x=6\)
Chọn A.
Phương trình
Lấy logarit cơ số 3 hai vế của (*), ta được
Suy ra
Cách nhanh nhất có loại câu này khỏi mục "chưa ai trả lời"
2.
c) /x-2/=9+15=24
x=26
hoạc
x=-22
\(5^{-1}.25^x=125\)
\(-5.25^x=125\)
\(25^x=125:\left(-5\right)\)
\(25^x=-25\)
\(\Rightarrow x=-1\)