Tìm x biết
( 3 x X - 16 ) x 343 - 2 x 2401
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#)Giải :
\(123-5\left(x+4\right)=38\)
\(\Leftrightarrow5\left(x+4\right)=85\)
\(\Leftrightarrow x+4=17\)
\(\Leftrightarrow x=13\)
\(\left(3x-16\right)343=2.2401\)
\(\Leftrightarrow\left(3x-16\right)343=4802\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
\(\left(2600+6400\right)-3x=1200\)
\(\Leftrightarrow9000-3x=1200\)
\(\Leftrightarrow3x=7800\)
\(\Leftrightarrow x=2600\)
P/s : Câu cuối cùng lỗi chỗ b4 ph k ? hay là tìm cả b và x thế ???
\(123-5\left(x+4\right)=38\)
\(=>123-5x-20=38\)
\(=>-5x=-123+20+38\)
\(=>-5x=-65\)
\(=>x=13\)
b, \(\left(3x-16\right)343=2.2401\)
=>1029x -5488= 4802
=> 1029x=5488+4802
=> 1029x= 10290
=> x=10
c, (2600+6400)-3x=1200
=> 9000-3x=1200
=> -3x= -9000+1200
=> -3x= -7800
=> x=2600
d,
k hiểu đề
3:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)
\(A=3^x+3^{x+1}+...+3^{x+100}\)
=>\(3A=3^{x+1}+3^{x+2}+...+3^{x+101}\)
=>\(2A=3^{x+101}-3^x\)
=>\(A=\dfrac{3^{x+101}-3^x}{2}\)
=>\(3^{x+101}-3^x=3^{105}-3^4\)
=>x=4
Bài 1 :
a) 72x-1 = 343
=> 72x-1 = 73
=> 2x - 1 = 3 => 2x = 4 => x = 2
b) (7x - 11)3 = 25.32 + 200
=> (7x - 11)3 = 32.9 + 200
=> (7x - 11)3 = 488
xem kĩ lại đề này :vvv
c) 174 - (2x - 1)2 = 53
=> (2x - 1)2 = 174 - 53
=> (2x - 1)2 = 174 - 125 = 49
=> (2x - 1)2 = (\(\pm\)7)2
=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)
Bài 2 :
a) x5 = 32 => x5 = 25 => x = 2
b) (x + 2)3 = 27
=> (x + 2)3 = 33
=> x + 2 = 3 => x = 3 - 2 = 1
c) (x - 1)4 = 16
=> (x - 1)4 = 24
=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)
d) (x - 1)8 = (x - 1)6
=> (x - 1)8 - (x - 1)6 = 0
=> (x - 1)6 [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)
+) x - 1 = 1 => x = 2 ( tm)
+) x - 1 = -1 => x = 0 ( tm)
Vậy x = 1,x = 2,x = 0
a) \(64^x:16^x=256\)
\(\Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=2^8\)
\(\Rightarrow2^{6x}:2^{4x}=2^8\)
\(\Rightarrow2^{6x-4x}=2^8\)
\(\Rightarrow2^{2x}=2^8\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
b) \(\dfrac{-2401}{7^x}=-7\)
\(\Rightarrow\dfrac{-7^4}{7^x}=-7\)
\(\Rightarrow-7^{4-x}=-7\)
\(\Rightarrow7^{4-x}=7\)
\(\Rightarrow4-x=1\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
c) \(\dfrac{64}{\left(-4\right)^x}=-256\)
\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)
\(\Rightarrow\left(-4\right)^x=-4\)
\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)
\(\Rightarrow x=1\)
\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)
\(b,\dfrac{-2401}{7^x}=-7\)
\(\Rightarrow7^x=-2401:\left(-7\right)\)
\(\Rightarrow7^x=343\)
\(\Rightarrow7^x=7^3\)
\(\Rightarrow x=3\)
\(c,\dfrac{64}{\left(-4\right)^x}=-256\)
\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)
\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)
\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)
\(\Rightarrow x=-1\)
#\(Toru\)
A, (2x+1)3=343
=> (2x+1)3=73
=> 2x + 1 = 7
=> 2x = 6
=> x = 3
B, 2x+2x+3=144
=> 2x+2x . 23 =144
=> 2x ( 1 + 23 ) =144
=> 2x ( 1 + 8 ) =144
=> 2x . 9 =144
=> 2 x = 16
=> 2 x = 2 4
=> x = 4
C, 3x+3x+2=2430
=> 3x+3x . 32 =2430
=> 3x . ( 1 + 32 ) =2430
=> 3x . ( 1 + 9 ) =2430
=> 3x . 10 =2430
=> 3x = 243
=> 3x = 3 5
=> x = 5