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bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
a/ \(3^{x+1}=9^x=3^{2x}\Rightarrow x+1=2x\Leftrightarrow x=1\)
b/ \(2^{3x+2}=4^{x+5}=2^{2x+10}\Rightarrow3x+2=2x+10\Leftrightarrow x=8\)
c/ \(3^{2x-1}=243=3^5\Rightarrow2x-1=5\Leftrightarrow x=3\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a) x + 1 = 8 => x=7
b) x+1=4 => x=3
c) x+1=2 => x=1
d) 3x + 2 =2(x+ 5 )
3x - 2x = -2 + 10
x = 8
e) 2x - 1 = 5 (vì 243=35)
=> x = 3
f) x - 1/2 = 1/3
=> x = 5/6
Chỗ nào k hiểu nói mình nhé
a/ (x+1)^2=64
=>(x+1)^2= 8^2
=>x+1 = 8
=>x = 8-1 = 7
b/2^x+1 = 16
=> 2^ x+1 = 2^4
=> x+1 = 4
=> x = 3
e/3^2x-1= 243
=>3^2x-1 = 3^5
=>2x-1 = 5
=>2x = 6
=>x =3
BT1: \(\left(3^2\right)^2-\left(-2^3\right)^2-\left(-5^2\right)^2=81-64-625=-608\)
BT2: a, \(\dfrac{1}{9}.27^x=3^x\)
\(3^{3x-2}=3^x\)
\(\Rightarrow3x-2=x\Rightarrow x=\dfrac{1}{2}\)
b, \(3^{-2}.3^4.3^x=3^7\)
\(3^{2+x}=3^7\Rightarrow2+x=7\)
\(\Rightarrow x=5\)
c, \(2^{-1}.2^x+4.2^x=9.2^5\)
\(2^x\left(2^{-1}+4\right)=288\)
\(\Rightarrow2^x=288:4,5=64=2^6\)
\(\Rightarrow x=6\)
d, \(\left(2x-3\right)^2=16=4^2\)
\(\Rightarrow2x-3=4\Rightarrow x=\dfrac{7}{2}\)
e, \(\left(3x-2\right)^5=-243=-3^5\)
\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}.\)
BT1: \(a,3^2.\dfrac{1}{243}.81^2.\dfrac{1}{33}=3^2.3^{-5}.3^8.3^{-1}\dfrac{1}{11}\)
\(=3^4.\dfrac{1}{11}=\dfrac{81}{11}\)
b, \(\left(4.5^3\right):\left(2^3.\dfrac{1}{10}\right)=100.5.\dfrac{1}{8}.10=625\)
thui tyuwj lamf ddi
1) 3x + 1 = 9x
3x + 1 = 32x
x + 1 = 2x
1 = 2x - x
1 = x
=> x = 1
2) 23x + 2 = 4x + 5
23x + 2 = 22(x + 5)
3x + 2 = 2x + 10
3x = 2x + 10 - 2
3x = 2x + 8
3x - 2x = 8
x = 8
=> x = 8
3x+ 1 = 9x
=> 3x + 1= 32x
=> x + 1 = 2x
=> x - 2x = -1
=> x = 1
23x+ 2= 4x + 5
=> 23x + 2 = 22x + 10
=> 3x + 2 = 2x + 10
=> 3x - 2x = 10 - 2
=> x = 8
32x - 1 =243
=> 32x - 1 = 35
=> 2x - 1 = 5
=> 2x = 5 + 1
=> 2x = 6
=> x = 3
1)
a) \(0,25^x\cdot12^x=243\)
\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(38^y:19^y=512\)
\(\Leftrightarrow2y\cdot y=512\)
\(\Leftrightarrow2y^2=512\)
\(\Leftrightarrow y^2=256\)
\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)
Vậy \(y_1=-16;y_2=16\)
2)
a) \(3^x+3^{x+2}=2430\)
\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)
\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)
\(\Leftrightarrow10\cdot3^x=2430\)
\(\Leftrightarrow3^x=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(2^{x+3}-2^x=224\)
\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)
\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
3)
a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)
b) \(\left(x+0,7\right)^3=-27\)
\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x+\dfrac{3}{10}=-3\)
\(\Leftrightarrow x=-3-\dfrac{3}{10}\)
\(\Leftrightarrow x=-\dfrac{37}{10}\)
Vậy \(x=-\dfrac{37}{10}\)
4)
a) \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{2}{5}-3x=\pm\dfrac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)
b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(\Leftrightarrow2x-1=1\)
\(\Leftrightarrow2x=1+1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
1. a) \(0,25^x.12^x=243\)
\(\Rightarrow\left(0,25.12\right)^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
b) \(38^y:19^y=512\)
\(\Rightarrow\left(38:19\right)^y=512\)
\(\Rightarrow2^y=2^9\)
\(\Rightarrow y=9\)
Vậy \(y=9.\)
2) a) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x\left(1+9\right)=2430\)
\(\Rightarrow3^x=243=3^5\)
\(\Rightarrow x=5\)
Vậy x=5.
b) \(2^{x+3}-2^x=224\)
\(\Rightarrow2^x\left(8-1\right)=224\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
Vậy x=5.
Bài 3: dễ tự làm.
a) \(2x.\left(x-\frac{1}{7}\right)=0\)
b) \(2^x=16\)
c) \(3^{x+1}=9^x\)
d) \(2^{3x-1}=243\)
Giup mk nka
a) \(2x.\left(x-\frac{1}{7}\right)=0\)
=>2x=0 hoặc \(x-\frac{1}{7}=0\)
+)Nếu 2x=0
=>x=0
+)Nếu \(x-\frac{1}{7}=0\)
=>x=\(\frac{1}{7}\)
b)2x=16
=>2x=(-2)4 hoặc 2x=24
=>x=4
c)3x+1=9x
=>\(3^{x+1}=\left(3^2\right)^x\)
=>\(3^{x+1}=3^{2x}\)
=>x+1=2x
=>x+1=x+x
=>x=1
d)23x-1=243 (mình nghĩ chỗ này bạn nhầm đề rồi, 33x-1 chứ ko phải 23x-1 đâu, nếu là 23x-1 thì sẽ ko tìm được x thỏa mãn)
33x-1=243
=>33x-1=35
=>3x-1=5
=>3x=5=1=6
=>x=6:3=2
a.
\(3^{x+1}=9^x\)
\(3^{x+1}=\left(3^2\right)^x\)
\(3^{x+1}=3^{2x}\)
\(x+1=2x\)
\(2x-x=1\)
\(x=1\)
c.
\(3^{2x-1}=243\)
\(3^{2x-1}=3^5\)
\(2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=\frac{6}{2}\)
\(x=3\)
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