8 mũ x nhân 16-2x= 45
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21 tháng 9 2020
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 - 22 = 0
<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0
<=> ( x - 5 )( x - 1 ) = 0
<=> x = 5 hoặc x = 1
b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16
<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> x = 50/40 = 5/4
d) x3 - 9x2 + 27x - 27 = -8
<=> ( x - 3 )3 = -8
<=> ( x - 3 )3 = (-2)3
<=> x - 3 = -2
<=> x = 1
e) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
21 tháng 9 2020
a) \(\left(x-3\right)^2-4=0\)
=> \(\left(x-3\right)^2-2^2=0\)
=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)
=> \(\left(x-5\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)
=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)
=> \(4x^2+12x+9-4x^2+1=22\)
=> \(12x+9+1=22\)
=> \(12x+10=22\)
=> 12x = 12
=> x = 1
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)
=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)
=> \(16x^2-9-16x^2+40x-25=16\)
=> \(-9+40x-25=16\)
=> \(40x=16+25-\left(-9\right)=16+25+9=50\)
=> x = 50/40 = 5/4
d) \(x^3-9x^2+27x-27=-8\)
=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)
=> \(\left(x-3\right)^3=-8\)
=> \(\left(x-3\right)^3=\left(-2\right)^3\)
=> x - 3 = -2 => x = 1
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)
=> \(3x+1=2\)
=> \(3x=1\)=> x = 1/3
f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)
=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)
=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)
=> \(\left(12x+x\right)-8=5\)
=> 13x = 13
=> x = 1
T
29 tháng 9 2017
A=5^10 .9+5^10.7 / 5^9.2^4
A=5^10.(9+7) / 5^9.16
A=5^10.16 /5^9.16
A=5^9.5.16/5^9.16
=>A=5/1
=>A=1
B=2^10.55+2^10. 26 / 2^8 . 27
B=2^10.(55+26) / 2^8 .27
B=2^10.81 / 2^8 .27
B=2^8.2^2.81 / 2^8.27
=>B=2^2.81 / 27
=>B=4.81 / 27
=>B=4.3^3.3 / 3^3
=>B=4.3
=>B=12
Cn phần C thì....mik chưa rõ lém nhưng mak nếu ai cko mik là dzui lém òy......ủng hộ nhoa mina
8 tháng 7 2018
\(8^4.16^5\)
\(=\left(2^3\right)^4.\left(2^4\right)^5\)
\(=2^{12}.2^{20}=2^{32}\)
30 tháng 7 2021
6, \(x^2y+xy^2-4x-4y=xy\left(x+y\right)-4\left(x+y\right)=\left(xy-4\right)\left(x+y\right)\)
7, \(10ax-5ay-2x+y=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(2x-y\right)\)
8, xem lại đề bạn nhé
9, \(4x^2-y^2+8y-16=4x^2-\left(y^2-8y+16\right)=4x^2-\left(y-4\right)^2\)
\(=\left(2x-y+4\right)\left(2x+y-4\right)\)
QA
30 tháng 7 2021
Trả lời:
6, x2y + xy2 - 4x - 4y = ( x2y + xy2 ) - ( 4x + 4y ) = xy ( x + y ) - 4 ( x + y ) = ( x + y )( xy - 4 )
7, 10ax - 5ay - 2x + y = ( 10ax - 5ay ) - ( 2x - y ) = 5a ( 2x - y ) - ( 2x - y ) = ( 2x - y )( 5a - 1 )
8, Sửa đề: x3 - 2x2 + 2x - 4 = ( x3 - 2x2 ) + ( 2x - 4 ) = x2 ( x - 2 ) + 2 ( x - 2 ) = ( x - 2 )( x2 + 2 )
9, 4x2 - y2 + 8y - 16 = 4x2 - ( y2 - 8y + 16 ) = 4x2 - ( y - 4 )2 = ( 2x - y + 4 )( 2x + y - 4 )
\(8^x.16^{-2x}=4^5\)
\(\Leftrightarrow2^{3x}.2^{-8x}=2^{10}\)
\(\Leftrightarrow2^{-24x^2}=2^{10}\)
\(\Leftrightarrow-24x^2=10\)
\(\Leftrightarrow x^2=\frac{10}{-24}\)
Mà \(x^2\ge0\)nên pt vô nghiệm