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(2x - 6)5 = (2x - 6)2
=> (2x - 6)5 - (2x - 6)2 = 0
=> (2x - 6)2.[(2x - 6)3 - 1] = 0
=> \(\orbr{\begin{cases}\left(2x-6\right)^2=0\\\left(2x-6\right)^3-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=6\\2x-6=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\2x=7\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=\frac{7}{2}\left(ktm\right)\end{cases}}\)
33x - 4 - x0 = 8
=> 33x - 4 - 1 = 8
=> 33x - 4 = 8 +1
=> 33x - 4 = 9
=> 33x - 4= 32
=> 3x - 4 = 2
=> 3x = 2 + 4
=> 3x = 6
=> x = 6 : 3 = 2
a) \(4^x=2^{x+1}\)
\(2^{2x}=2^{x+1}\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow2x-x=1\)
\(\Rightarrow x=1\)
b) \(16=\left(x-1\right)^4\)
\(2^4=\left(x-1\right)^4\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x=3\)
c) \(x^{10}=1^x\)
\(x^{10}=1\)
\(x^{10}=1^{10}\)
\(\Rightarrow x=1\)
d) \(x^{10}=x\)
\(x^{10}-x=0\)
\(x\left(x^9-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=\pm1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{15}{2}\\x=\left\{8;7\right\}\end{cases}}\)
\(A,4^X=2^{X+1}\)
\(\left(2^2\right)^X=2^{X+1}\)
\(\Rightarrow2^{2X}=2^{X+1}\)
\(\Rightarrow2X=X+1\)
\(\Rightarrow2X-X=1\Leftrightarrow X=1\)
\(B,16=\left(x-1\right)^4\)
\(\Rightarrow x-1=\hept{\begin{cases}2\\-2\end{cases}}\)
\(\Rightarrow x=\hept{\begin{cases}3\\-1\end{cases}}\)
Bài 1 :
\(2^x.8=512\)
\(2^x=512:8\)
\(2^x=64\)
\(2^x=2^6\)
\(\Rightarrow x=6\)
\(b,\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(c,x^{20}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(d,\left(x-3\right)^{10}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
Câu 1.
C = 5 + 42 + 43 + ... + 42020
a) Xét A = 42 + 43 + ... + 42020
=> 4A = 43 + 44 + ... + 42021
=> 4A - A = 3A
= 43 + 44 + ... + 42021 - ( 42 + 43 + ... + 42020 )
= 43 + 44 + ... + 42021 - 42 - 43 - ... - 42020
= 42021 - 42
=> A = \(\frac{4^{2021}-4^2}{3}\)
Thế vào C ta được : \(C=5+\frac{4^{2021}-4^2}{3}=\frac{15}{3}+\frac{4^{2021}-4^2}{3}=\frac{4^{2021}+15-16}{3}=\frac{4^{2021}-1}{3}\)
b) D = 42021 => \(\frac{D}{3}=\frac{4^{2021}}{3}\)
Vì 42021 - 1 < 42021 => \(\frac{4^{2021}-1}{3}< \frac{4^{2021}}{3}\)
=> C < D/3
c) Dùng kết quả ý a) ta được :
3C + 1 = 42x-6
<=> \(3\cdot\frac{4^{2021}-1}{3}+1=4^{2x-6}\)
<=> 42021 - 1 + 1 = 42x-6
<=> 42021 = 42x-6
<=> 2021 = 2x - 6
<=> 2x = 2027
<=> x = 2027/2
Câu 2.
( x - 1 )( 4 + 22 + 23 + ... + 220 ) = 222 - 221
Xét A = 22 + 23 + ... + 220
=> 2A = 23 + 24 + ... + 221
=> A = 2A - A
= 23 + 24 + ... + 221 - ( 22 + 23 + ... + 220 )
= 23 + 24 + ... + 221 - 22 - 23 - ... - 220
= 221 - 4
Thế vô đề bài ta được
( x - 1 )( 4 + 221 - 4 ) = 222 - 221
<=> ( x - 1 ).221 = 221( 2 - 1 )
<=> x - 1 = 1
<=> x = 2
a) 52 . x = 62 + 82
\(5^2\cdot x=36+64\)
\(5^2\cdot x=100\)
\(x=100\div5^2\)
\(x=100\div25\)
\(x=4\)
b) ( 22 + 42 ) . x + 24 . 5 . x = 102
\(\left(4+16\right)\cdot x+16\cdot5\cdot x=100\)
\(x\cdot\left(20+80\right)=100\)
\(x\cdot100=100\)
\(x=100\div100\)
\(x=1\)
c ) 24 . x = 26
\(x=2^6\div2^4\)
\(x=2^{6-4}\)
\(x=2^2\)
\(x=4\)
d) 33 . x + 23 . x = 102
\(x\cdot\left(23+27\right)=100\)
\(x\cdot50=100\)
\(x=100\div50\)
\(x=2\)
e) 78 . x = 710
\(x=7^{10}\div7^8\)
\(x=7^{10-8}\)
\(x=7^2\)
\(x=49\)
13/ => 10 + 2x = 42 = 16
=> 2x = 6
=> x = 3
14/ => 52x : 53 - 50 = 75
=> 52x : 53 = 125 = 53
=> 52x = 56
=> 2x = 6
=> x = 3
15/ => (26 - 3x) : 5 = 4
=> 26 - 3x = 20
=> 3x = 6
=> x = 2
16/ => x - 17 = -25
=> x = -8
a) 52.x = 62 + 82
=> 25 .x = 36 + 64
=> 25.x = 100
=> x = 100 : 25
=> x = 4
b) (22 + 42).x + 24 . 5x = 100
=> (4 + 16).x + 16.5x = 100
=> 20x + 80x = 100
=> 100x = 100
=> x = 100 : 100 = 1
c) 24 : x = 26
=> x = 24 : 26
=> x = 2-2 = 1/4
d) 33x + 23x = 102
=> 27x + 8x = 100
=> 35x = 100
=> x = 100 : 35
=> x = 20/7
\(10:2x=16\\ 2x=\dfrac{5}{8}\\ x=\dfrac{5}{16}\)
10 + 2x=4^5 : 4^3
10 + 2x =4^2
10 + 2x =16
2x=16 - 10
2x=6
x=6:2
x=3