tìm x thuộc N biết
2^x+2+2^x=144
(2x+1)^3=243
tìm x
3^x-1=1/243
81^-2x.27^x=9^5
2^x+2^x+3=144
3^ x -1 = 1/243
3^x =1/243 +1
3^x = 244 / 243
Ta thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn
81^-2x . 27^x =9^5
81^-2 . 81^x . 27^x =9^5
1/9^4 . (81.27)^x =9 ^5
3^6x = 9^5 : 1/9^4
3^6x = 9^9
3^6x = 3^18
=> 6x =18
x=3
2^x +2^x +3 =144
2.(2^x) =141
2^x+1 = 141
Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn
1)tìm x biết:
a)(x-2)^2=1
b)(2x-1)^3=-8
c)(x+1/2)^2=1/16
d)(x-2)^3=-27
e(2x-3)^2=25
g)3^x-1=1/243
h)2^x+2^x+3=144
k)81^-2x*27^x=95
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
g.\(3^{x-1}\)=\(\frac{1}{243}\)
\(3^{x-1}\)=\(\frac{1}{3^5}\)
\(3^{x-1}\)=\(3^{-5}\)
x-1=-5
x=-4
tìm x, biết
2.(3/5)^x + 1/5 =2/25^3
1)tìm x biết:
a)(x-2)^2=1
b)(2x-1)^3=-8
c)(x+1/2)^2=1/16
d)(x-2)^3=-27
e(2x-3)^2=25
g)3^x-1=1/243
h)2^x+2^x+3=144
k)81^-2x*27^x=95
a) (X - 2)\(^2\) = 1 <=> X - 2 = \(\sqrt{1}\) <=> X = 1 + 2 <=> X = 3
Tìm x thuộc n biết 3 mũ 2 x +1 = 243
\(3^{2x+1}=243\)
\(\Rightarrow3^{2x+1}=3^5\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
\(3^2x+1=243\)
\(9x=243-1\)
\(9x=242\)
\(x=242:9\)
\(\Rightarrow x=\dfrac{242}{9}\)
tìm x thuộc N biết (2x+1)3=243
tìm x biết:
a, (x-2)3: 3 = -9
b, 3x-1= 1/243
c, (2x-3)2 = 1/4
d,, 2x+ 2x-3 =144
\(a.\) \(\left(x-2\right)^3:3=-9\)
\(\Leftrightarrow\left(x-2\right)^3=-9.3\)
\(\Leftrightarrow\left(x-2\right)^3=-27\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-2=-3\)
\(\Leftrightarrow x=-3+2\)
\(\Leftrightarrow x=-1\)
\(b.\) \(3^{x-1}=\frac{1}{243}\)
\(\Leftrightarrow3^x:3=\frac{1}{243}\)
\(\Leftrightarrow3^x=\frac{1}{243}.3\)
\(\Leftrightarrow3^x=\frac{1}{81}\)
\(\Leftrightarrow3^x=3^{-4}\)
\(\Leftrightarrow x=-4\)
\(c.\) \(\left(2x-3\right)^2=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-3\right)^2=\frac{1}{2}^2\\\left(2x-3\right)^2=\left(-\frac{1}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{2}+3\\2x=-\frac{1}{2}+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{2}\\2x=\frac{5}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}:2=\frac{7}{4}\\x=\frac{5}{2}:2=\frac{5}{4}\end{cases}}\)
\(d.\) Thiếu đề rồi bn !!!
Câu d bn vt nhầm đề đúng ko ???
\(d.\) \(2^x+2^{x-3}=144\)
\(\Leftrightarrow2^x.1+2^x:2^3=144\)
\(\Leftrightarrow2^x.1+2^x.\frac{1}{8}=144\)
\(\Leftrightarrow2^x.\left(1+\frac{1}{8}\right)=144\)
\(\Leftrightarrow2^x.\frac{9}{8}=144\)
\(\Leftrightarrow2^x=144:\frac{9}{8}\)
\(\Leftrightarrow2^x=128\)
\(\Leftrightarrow2^x=2^7\)
\(\Leftrightarrow x=7\)
bài 1 tìm x thuộc n
a) 1/9 * 3 mũ 4 * 3 mũ n = 3 mũ 7
b) 2.3 mũ x = 26. 2 mũ 2 +2.30
c) 10 mũ x -3 =100
d) (x-3) mũ 3 =144
e) (2x+1) mũ 3 =125
Nhờ mọi người giúp mình vs!
Tìm x thuộc Q khi:
a)3x-1 =243
b)\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
c) 3x+3x+2=270
d) \(\left(x-\dfrac{5}{2}\right)^2=144\)
e) \(\left(2x+1\right)^3=\dfrac{215}{27}\)
f) (2x-1)5 = -243
h) (27-1)2 = (2x-1)2
i) (2x+1)2 + (x+1)10 = 0
k) \(\dfrac{x-2}{5}=\dfrac{X+3}{2}\)
l) \(\dfrac{2}{x}=\dfrac{x}{50}\)
m) \(\dfrac{\left(x-3\right)}{4}=\dfrac{4}{x+3}\)
n) \(|x+\dfrac{3}{5}|-|x-\dfrac{7}{3}|\)
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.