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a) 2x+2 - 2x= 192
2x . 22 - 2x . 1 = 192
2x . ( 22 - 1) = 192
2x . 3 = 192
2x = 192 : 3
2x = 64 = 26
=> x = 6
a. (x-2)3=27
<=> (x-2)3=33
=> x-2=3 =>x=5
c.(2x-3)2=25
<=> (2x-3)2=52
=>2x-3=5 => 2x=8 =>x=4
a) ( x-2)3 = 27
(x-2)3 = 33
x-2 = 3
x = 5
b) 2x + 2x+3= 144
2x ( 1 + 23 ) = 144
2x . 9 = 144
2x = 16
2x = 24
x = 4
c) ( 2x-3)2= 25
(2x-3)2 = 52
2x-3 = 5
2x = 8
x = 4
d) ( 2x-1)10= ( 2x-1) 100
Dễ thấy 2x - 1 = 0 vì 010 = 0100
Vậy x = 0,5
Nếu 2x - 1 \(\ne\)0
( 2x - 1 )10 : ( 2x - 1 )10 = ( 2x - 1 )100 : ( 2x - 1 )10
( 2x - 1 )90 = 1
Vậy 2x - 1 = 1 hoặc - 1.
Vậy x = 1 ; 0 ; 0,5.
Nhiều câu quá >.<
a/ \(2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20.\)
\(2x^2+10x=x^2+6x+9+x^2-2x+1+20.\)
\(10x=4x+30\)
\(6x=30\Rightarrow x=5\)
các câu còn lại tương tự
\(a,2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)
\(\Leftrightarrow2x^2+10x=x^2+6x+9+x^2-2x+1+20\)
\(\Leftrightarrow2x^2+10x=2x^2+4x+30\)
\(\Leftrightarrow2x^2+10x-2x^2-4x=30\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
Vậy ...........
\(b,\left(2x-2\right)^2=\left(x+1\right)^2+3\left(x-2\right)\left(x+5\right)\)
\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3x^2+15x-6x-30\)
\(\Leftrightarrow4x^2-8x+4=4x^2+11x-29\)
\(\Leftrightarrow4x^2-8x-4x^2-11x=-29-4\)
\(\Leftrightarrow-19x=-33\)
\(\Leftrightarrow x=\frac{33}{19}\)
Vậy...........
\(c,\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)
\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2x^2+2x-4x-4+38\)
\(\Leftrightarrow2x^2+4x+10=2x^2-2x+34\)
\(\Leftrightarrow2x^2+4x-2x^2+2x=34-10\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy.............
\(d,\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-18\)
\(\Leftrightarrow x^3+6x+12x+8-\left(x^3-6x+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x+12x+8-x^3+6x-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy............
a. => \(2^{6+x}=2^{10}\)
=> 6+x=10
=> x=10-6
Vậy x=4.
b. => \(7^{3x-1}:7^2=7^6\)
=> 73x-1-2=76
=> 73x-3=76
=> 3x-3=6
=> 3x=6+3
=> 3x=9
Vậy x=3.
c. =>\(7^{5x-1}-25=24\)
=>75x-1=24+25
=>75x-1=49
=>75x-1=72
=>5x-1=2
=>5x=3
Vậy x=\(\frac{3}{5}\).
d. => \(10^{x-3}=10^0\)
=>x-3=0
Vậy x=3.
e. => 2x=10
=> x=10:2
Vậy x=5.