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gọi A = {1;3;5;...;2x+1 / x thuộc N}
=> Số số hạng của A là:
(2x + 1 - 1) : 2 = x.
=> (1 + 2x + 1)x : 2 = 224
<=> 2(x+1)x:2=224
<=> x(x+1)=224
Mặt khác: x và x+1 là 2 số tự nhiên liên tiếp và tích của chúng chỉ tận cùng = 0;2;6
=> o tồn tại stn x thỏa mãn đề bài.
Vậy x thuộc tập rỗng
\(2x-124=x+224\\ \Rightarrow2x+x=224+124\\ \Rightarrow3x=348\\ \Rightarrow x=348:3\\ \Rightarrow x=116\)
2\(x\) - 124 = \(x\) + 224
2\(x\) - \(x\) = 224 + 124
\(x\) = 348
2 mũ x nhân 7=224 (3x+5) mũ 2=289 phần c mình chịu T-T
2 mũ x=224:7 (3x+5) mũ 2=17 mũ 2
2 mũ x=32 3x+5=17
2 mũ 5=32 3x=17-2
=>x=5 3x=15
x=15:3
x=5
a) \(\left(x-1\right)^3=125\)
\(\Leftrightarrow\left(x-1\right)^3=5^3\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
b) \(2^{x+2}-2^x=96\)
\(\Leftrightarrow\left(2^2-1\right)\cdot2^x=96\)
\(\Leftrightarrow\left(4-1\right)\cdot2^x=96\)
\(\Leftrightarrow3\cdot2^x=96\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
c) \(\left(2x+1\right)^3=343\)
\(\Leftrightarrow\left(2x+1\right)^3=7^3\)
\(\Leftrightarrow2x+1=7\)
\(\Leftrightarrow2x=7-1\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
d) \(720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=2^3\cdot5\left(đk:x\ne23\right)\)
\(\Leftrightarrow720:\left(41-2x+5\right)=8\cdot5\)
\(\Leftrightarrow720:\left(46-2x\right)=40\)
\(\Leftrightarrow\dfrac{720}{46-2x}=40\)
\(\Leftrightarrow\dfrac{720}{2\left(23-x\right)}=40\)
\(\Leftrightarrow\dfrac{360}{23-x}=40\)
\(\Leftrightarrow360=40\left(23-x\right)\)
\(\Leftrightarrow9=23-x\)
\(\Leftrightarrow x=23-9\)
\(\Leftrightarrow x=14\left(đk:x\ne23\right)\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
e) \(2^x\cdot7=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
f) \(\left(3x+5\right)^2=289\)
\(\Leftrightarrow3x+5=\pm17\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=17\\3x+5=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{22}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{22}{3};x_2=4\)
a)\(\left(x-1\right)^3=125\Leftrightarrow\left(x-1\right)^3=5^3\Leftrightarrow x-1=5\Leftrightarrow x=6\)b)\(2^{x+2}-2^x=96\Leftrightarrow2^x.2^2-2^x=96\Leftrightarrow2^x\left(2^2-1\right)=96\Leftrightarrow2^x.3=96\Leftrightarrow2^x=32\Leftrightarrow x=5\)c)\(\left(2x-1\right)^3=343\Leftrightarrow\left(2x-1\right)^3=7^3\Leftrightarrow2x-1=7\Rightarrow2x=8\Rightarrow x=4\)d)\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-\left(2x-5\right)\right]=40\Leftrightarrow\left[41-\left(2x-5\right)\right]=720:40=18\)
\(\Leftrightarrow41-2x+5=18\Leftrightarrow36-2x=18\Leftrightarrow2x=18\Leftrightarrow x=9\)
e)\(2^x.7=224\Leftrightarrow2^x=224:7=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
f) \(\left(3x+5\right)^2=289\Leftrightarrow\left(3x+5\right)=17^2\Leftrightarrow3x+5=17\Leftrightarrow3x=12\Leftrightarrow x=4\)
a,x2.x3=25
=>x5=25
=>x=2
b,x+18=5.4^2
=>x=5.16-18
=>x=62
c,x.(x^2)^3=x^5
=>x.x5=x5
=>x=0,1
d,2x.7=224
2x=32
=>2x=25
=>x=5
e,(3x+5)2=289
=>(3x+5)2=172
=>3x+5=17
=>3x=12
=>x=4
g,32x+1.11=2673
=>32x=243
=>32x=35
=>x=\(\frac{5}{2}\)
a) \(15+2\left|x\right|=-3\\ \\ < =>2\left|x\right|=15-\left(-3\right)\\ < =>2\left|x\right|=18\\ =>\left|x\right|=\frac{18}{2}=9\\ =>x=9hoặcx=-9\)
b) \(\left|x-2\right|=7\\ < =>x-2=7hoặcx-2=-7\\ =>x=9hoặcx=-5\)
c) \(100-4.x^2=224\\ < =>4.x^2=100-224=-124\\ < =>x^2=-\frac{124}{4}=-31\\ Mà:x^2\ge0\\ =>xkhôngcógiátrịnàothoảmãn\)
d)\(2x-\frac{9}{240}=\frac{39}{80}\\ < =>2x-\frac{3}{80}=\frac{39}{80}\\ =>2x=\frac{39}{80}+\frac{3}{80}=\frac{21}{40}\\ =>x=\frac{\frac{21}{40}}{2}=\frac{21}{80}\)
a/ 2x . 7 = 224
2x = 224 : 7
2x = 32
2x = 25
x = 5.
b/ 32x + 1 . 11 = 2673
32x + 1 = 2673 : 11
32x + 1 = 243
32x + 1 = 35
32x = 35 - 1
32x = 34
2x = 4
x = 4 : 2
x = 2.
c/ (3x + 5)2 = 289
(3x + 5)2 = 172
3x + 5 = 17
3x = 17 - 5
3x = 12
x = 12 : 3
x = 4.
d/ x . (x2)3 = x5
x1 . x5 = x5
x6 = x5
=> x = 0 hoặc x = 1
a) 2x x 7=224
2x=224:7
2x=32
2x=25
=> x=5
Vậy x=5
b) 32x+1 x 11=2673
32x+1=2673:11
32x+1=243
32x+1=35
=> 2x+1=5
x=(5-1):2
x=2
Vậy x=2
c) (3*x+5)2=289
(3*x+5)2=172
=> 3*x+5=17
x=(17-5):3
x=4
Vậy x=4
d) x.(x2)3=x5
x.x6=x5
x=x5:x6
x=x-1
a) \(x-\frac{1}{12}+x-\frac{1}{20}+x-\frac{1}{30}+x-\frac{1}{42}+x-\frac{1}{56}+x-\frac{1}{72}=224\)
\(\left(x+x+x+x+x+x\right)-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=224\)
\(6x-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=224\)
\(6x-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=224\)
\(6x-\left(\frac{1}{3}-\frac{1}{9}\right)=224\)
\(6x-\frac{2}{9}=224\)
\(6x=224+\frac{2}{9}\)
\(6x=\frac{2018}{9}\)
\(\Rightarrow x=\frac{2018}{9}:6=\frac{1009}{27}\)
b) ( 2x - 1 ) 2 = \(\frac{1}{4}\)
( 2x - 1 ) 2 = \(\left(\frac{1}{2}\right)^2\)
\(\Rightarrow\)2x - 1 = \(\frac{1}{2}\)
\(\Rightarrow\)2x = \(\frac{1}{2}+1\)
\(\Rightarrow\)2x = \(\frac{3}{2}\)
\(\Rightarrow\)x = \(\frac{3}{2}:2\)
\(\Rightarrow\)x = \(\frac{3}{4}\)
2.
Ta có : 3300 = ( 33 ) 100 = 27100
5200 = ( 52 ) 100 = 25100
Vì 27100 > 25100 nên 3300 > 5200
3.
150 - ( 100 - 99 + 98 - 97 + 96 - 95 + ... + 4 - 3 + 2 - 1 )
= 150 - [ (100 - 99 ) + ( 98 - 97 ) + ( 96 - 95 ) + ... + ( 4 - 3 ) + ( 2 - 1 ) ]
= 150 - ( 1 + 1 + 1 + ... + 1 + 1 )
= 150 - 50
= 100
4.
ta có :
9x + 5y + 4 .( 2x + 3y )
= 9x + 5y + 8x + 12y
= ( 9x + 8x ) + ( 5y + 12y )
= 17x + 17y
= 17 ( x + y ) \(⋮\)17
Vì 9x + 5y \(⋮\)17 \(\Rightarrow\)4 . ( 2x + 3y ) \(⋮\)17
Mà ( 4 ; 17 ) = 1
\(\Rightarrow\)2x + 3y \(⋮\)17
bài 1
a) \(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=224\)
\(\left(x-1\right).\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=224\)
\(\left(x-1\right).\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=224\)
\(\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=24\)
\(\left(x-1\right).\left(\frac{1}{3}-\frac{1}{9}\right)=\left(x-1\right)\cdot\frac{2}{9}=224\)
\(\Rightarrow\left(x-1\right)=224:\frac{2}{9}=1008\Rightarrow x=1008+1=1009\)
\(2^{x+3}-2^x=224\)
=>\(2^x\cdot8-2^x=224\)
=>\(7\cdot2^x=7\cdot32\)
=>\(2^x=32=2^5\)
=>x=5
2x+3 - 2x = 224
2x+3 - 2x = 28- 25
=> x+3 - x = 8 - 5
3 = 3
=> pt luôn bằng 3 với mọi x