(3x-2)5 = 25 . 125
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125-5(3x-1)=25
5(3x-1)=125-25
5(3x-1)=100
3x-1=100:5
3x-1=20
3x =20+1
3x = 21
x=21:3
x= 7
Vậy x = 7
125-5 3x = 25+1
125-5 3x = 26
120 3x = 26
3x = 120 . 26
3x = 3,120 : 3
x = 1,040
Làm đúng ko ?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
125 - 5.(3x - 1) = 25
5.(3x - 1) = 125 - 25
5.(3x - 1) = 100
3x - 1 = 100 : 5
3x - 1 = 20
3x = 20 + 1
3x = 21
x = 21 : 3
x = 7
Vậy x = 7
a ) 126 - 2x = 12 - ( -28 )
126 - 2x = 40
2x = 126 - 40
2x = 86
x = 86 : 2
x = 43
b ) 25 - 5( x + 1 ) = 30 - 125
25 - ( 5x + 5 ) = -95
25 - 5x - 5 = -95
20 - 5x = -95
5x = 20 - ( - 95 )
5x = 115
x = 115 : 5
x = 23
c ) I 3x - 1 I - 25 = 73
I 3x - 1 I = 73 + 25
I 3x - 1 I = 98
=> 3x - 1 = 98 hoặc 3x - 1 = -98
=> 3x = 98 + 1 hoặc 3x = -98 + 1
=> 3x = 99 hoặc 3x = -97
=> x = 99 : 3 hoặc x = -97 : 3
=> x = 33 hoặc x = -93/3
d ) 2( x + 2 ) - 6 = 120 - 4x
2x + 4 - 6 = 120 - 4x
2x - 2 = 120 - 4x
2x + 4x = 120 + 2
6x = 122
x = 122 : 6
x = 61/3
( 3x - 1 )^2 = 25 = (+-5)^2
+) 3x - 1 = 5
3x = 6
x = 2
+) 3x - 1 = -5
3x = -4
x = -4/3
Vậy,.........
Các câu còn lại tương tự
Bài 9:
a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)
\(=3xy-y^2\)
\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)
b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{31}{2}\)
Bài 7:
a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)
b) \(93\cdot107=100^2-7^2=10000-49=9951\)
c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)
d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)
e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1=1\)
f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
a) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
b) \(3^x+25=26\times2^2+2\times3^0\)
\(3^x+25=26\times4+2\times1\)
\(3^x+25=106\)
\(3^x=106-25\)
\(3^x=81\)
\(3^x=3^4\)
\(x=4\)
(2x+1)3 = 125
a)<=> (2x+1)3 = 53
<=> 2x+1 = 5
<=> 2x = 4
<=> x = 2
3^x+25=26 . 2^2 + 2. 3^0
b)3^x+25=104 +2
3^x+25=106
3^x=106+25
3^x=81=3^4
=> x=4
( 3x - 2 )5 = 52 . 53
( 3x -2 )5 = 55
=> 3x-2=5
3x=5+2
3x=7
x=\(\frac{7}{3}\)
(3x-2)5=25×125
(3x-2)5=52×53
(3x-2)5=55
=>3x-2=5
3x =7
x =7÷3
x = 7/3