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Bài 1 :
A = 1 + 2 + 22 + ... + 211
A = ( 1 + 2 ) + ( 22 + 23 ) + ... + ( 210 + 211 )
A = 3 + 22(1+2) + ... + 210(1+2)
A = 1.3 + 22.3 + ... + 210.3
A = 3.(1+22+...+210) chia hết cho 3
Bài 2 :
2.52 + 3:710 - 54:33
= 2.25 + 3:1 - 54:27
= 50 + 3 - 2
= 49
Bài 3 :
a) ( 2x - 6 ) . 47 = 49
2x - 6 = 42 = 16
2x = 16
=> x = 8
b) ( 27x + 6 ) : 3 - 11 = 9
( 27x + 6 ) : 3 = 20
27x + 6 = 60
27x = 54
=> x = 2
c) 740 : ( x + 10 ) = 102 - 2.13
740 : ( x + 10 ) = 74
x + 10 = 10
=> x = 0
d) ( 15 - 6x ) . 35 = 36
15 - 6x = 3
6x = 12
=> x = 2
Bài 4 :
Ta có : ab + ba = ( 10a + b ) + ( 10b + a ) = ( 10a + a ) + ( 10b + b ) = 11a + 11a = 11.(a+b) chia hết cho 11
Bài 1 :
A = 1 + 2 + 22 + ... + 211
A = ( 1 + 2 ) + ( 22 + 23 ) + ... + ( 210 + 211 )
A = 3 + 22(1+2) + ... + 210(1+2)
A = 1.3 + 22.3 + ... + 210.3A = 3.(1+22+...+210) chia hết cho 3
Bài 2 :
2.52 + 3:710 - 54:33
= 2.25 + 3:1 - 54:27
= 50 + 3 - 2= 49
Bài 3 :
a) ( 2x - 6 ) . 47 = 49
2x - 6 = 42 = 16
2x = 16
=> x = 8
b) ( 27x + 6 ) : 3 - 11 = 9
( 27x + 6 ) : 3 = 20
27x + 6 = 60
27x = 54
=> x = 2
c) 740 : ( x + 10 ) = 102 - 2.13
740 : ( x + 10 ) = 74
x + 10 = 10
=> x = 0
d) ( 15 - 6x ) . 35 = 36
15 - 6x = 3
6x = 12
=> x = 2
Bài 4 :
Ta có : ab + ba = ( 10a + b ) + ( 10b + a ) = ( 10a + a ) + ( 10b + b ) = 11a + 11a = 11.(a+b) chia hết cho 11
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
Câu 2:
a: \(x^{10}=1^x\)
\(\Leftrightarrow x^{10}=1\)
=>x=1 hoặc x=-1
b: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
hay \(x\in\left\{\dfrac{15}{2};8;7\right\}\)
c: \(x^{10}=x\)
\(\Leftrightarrow x\left(x^9-1\right)=0\)
=>x=0 hoặc x=1