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a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a: =>9x=5(1-4x)
=>9x=5-20x
=>29x=5
hay x=5/29
b: =>x/5=5/x
=>x2=25
=>x=5 hoặc x=-5
c: =>(x+2)3=1/343
=>x+2=1/7
hay x=-13/7
Câu 1:
a. $=-(35+65)+(22+88)=-100+110=110-100=10$
c. $=58(26+74)=58.100=5800$
b. $=29(38+63-1)=29.100=2900$
b) \(\Leftrightarrow18x-288=27.4+8.9-144\)
\(\Leftrightarrow18x=108+72-144+288=324\)
\(\Leftrightarrow x=18\)
a) 33x+1 . 5 = 10935
=> 33x . 3 . 5 = 10935
=> 27x . 3 . 5 = 10935
=> 27x = 729
=> 27x = 272
=> x = 2
b) 25x + 1 . 3 = 6144
=> 25x . 2 . 3 = 6144
=> 32x . 6 = 6144
=> 32x = 1024
=> 32x = 322
=> x = 2
c) (2x + 3)4 = 625
=> (2x + 3)4 = (\(\pm\)5)4
=> \(\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
d) (3x + 1)3 = 343
=> (3x + 1)3 = 73
=> 3x + 1 = 7 => 3x = 6 => x = 2
a) \(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)
b) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+...+3^{101}\)
\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)
\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)
c) \(C=5+5^2+...+5^{30}\)
\(\Rightarrow5C=5^2+5^3+...+5^{31}\)
\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)
\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)
d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
a) 17 + x = -5 + 2x
=> x - 2x = -5 - 17
=> -x = -22
=> x = 22
b) 15 - 3x = -22 - 4x
=> -3x + 4x = -22 - 15
=> x = -37
c) 47 - (4x + 5) = -(3x - 2)
=> 47 - 4x - 5 = -3x + 2
=> -4x + 3x = 2 - 47 + 5
=> -x = -40
=> x = 40
d) -58 + (3 - 7x) = 15 - 8x
=> -58 + 3 - 7x = 15 - 8x
=> -7x + 8x =15 + 58 - 3
=> x = 70
a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)
=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)
=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)
b) \(3x^2-2=72\)=> 3x2 = 74 => x2 = 74/3 => x không thỏa mãn
c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)
=> \(\left(19x+50\right):14=9\)
=> \(19x+50=126\)
=> \(19x=76\)
=> x = 4
d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)
=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)
=> \(x\left(2+4+8+16+32\right)=343\)
=> x . 62 = 343
=> x = 343/62
a) (22-29)3=343
(4-29 )3 =343
( -25)3 = 343
( -25)3 = 73
Vậy x \(\in\varnothing\)
~HT~
\(c,3^x.3^2+2-3^x=297\)
\(3^x.\left(9+2\right)=297\)
\(3^x.11=297\)
\(3^x=297:11\)
\(3^x=27\)
\(< =>3^3=27=>x=3\)
còn phần a mk ko bt x là j nhé