HT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TH
3
K
31 tháng 8 2015
a) (2x + 3)2 = 9/121
Ta có: 9/121 = (3/11)2 = (-3/11)2
=> 2x + 3 thuộc {3/11; -3/11}
=> x thuộc {-15/11; -18/11}
b) (3x - 1)3 = -8/27 = (-2/3)3
=> 3x - 1 = -2/3
=> x = 1/9
7 tháng 12 2018
\(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Rightarrow\left(2x+3\right)^2=\hept{\begin{cases}\left(\frac{3}{11}\right)^2\\\left(\frac{-3}{-11}\right)^2\end{cases}}\)
\(\Rightarrow2x+3=\hept{\begin{cases}\frac{3}{11}\\\frac{-3}{-11}\end{cases}}\)
DT
2
N
30 tháng 10 2019
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
\(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\frac{2}{3}\)
\(3x=-\frac{2}{3}+1\)
\(3x=\frac{1}{3}\)
\(x=\frac{1}{3}:3\)
\(x=\frac{1}{3}.\frac{1}{3}\)
\(x=\frac{1}{9}\)
Vậy x = 1/9
c) |2x + 1| - 3 = 6
|2x + 1| = 6 + 3
|2x + 1| = 9
=> |2x + 1| ∈ {9;-9}
\(\Rightarrow\hept{\begin{cases}2x+1=9\\2x+1=-9\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=9-1\\2x=-9-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=8\\2x=-10\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=8:2=4\\x=-10:2=-5\end{cases}}\)
Vậy x ∈ {4;-5}
AH
Akai Haruma
Giáo viên
7 tháng 8 2019
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
AH
Akai Haruma
Giáo viên
7 tháng 8 2019
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
2 tháng 10 2020
a) ( x - 1/5 )2 = 0
<=> x - 1/5 = 0
<=> x = 1/5
b) ( x - 2 )2 = 1
<=> ( x - 2 )2 = ( ±1 )2
<=> x - 2 = 1 hoặc x - 2 = -1
<=> x = 3 hoặc x = 1
c) ( 2x - 1 )3 = -8
<=> ( 2x - 1 )3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = -1/2
d) ( x4 )2 = x12/x5
<=> x8 = x7
<=> x8 - x7 = 0
<=> x7( x - 1 ) = 0
<=> x7 = 0 hoặc x - 1 = 0
<=> x = 0 hoặc x = 1
e) x10 = 25x8
<=> x10 - 25x8 = 0
<=> x8( x2 - 25 ) = 0
<=> x8 = 0 hoặc x2 - 25 = 0
<=> x = 0 hoặc x = ±5
f) ( 2x + 3 )2 = 9/121
<=> ( 2x + 3 )2 = ( ±3/11 )2
<=> 2x + 3 = 3/11 hoặc 2x + 3 = -3/11
<=> x = -15/11 hoặc x = -18/11
2 tháng 10 2020
a) \(\left(x-\frac{1}{5}\right)^2=0\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3+8=0\)
\(\Leftrightarrow\left(2x-1+8\right)\left[\left(2x-1\right)^2-8\left(2x-1\right)+64\right]=0\)
\(\Leftrightarrow2x+7=0\)
\(\Leftrightarrow x=\frac{-7}{2}\)
d) ĐKXĐ : \(x\ne0\)
\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)
\(\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\)
\(\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=1\left(tm\right)\end{cases}\Leftrightarrow x=1}\)
e) ĐKXĐ : x khác 0
\(x^{10}=25x^8\)
\(\Leftrightarrow x^2=25\Leftrightarrow x=5\)
f) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Leftrightarrow\left(2x+3+\frac{3}{11}\right)\left(2x+3-\frac{3}{11}\right)=0\)
\(\Leftrightarrow\left(2x+\frac{36}{11}\right)\left(2x+\frac{30}{11}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-18}{11}\\x=-\frac{15}{11}\end{cases}}\)
14 tháng 7 2018
a) \(\left(2x+3\right)^2=\frac{9}{144}\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)
Vậy ...
b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)
Vậy ....
c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)
Vậy ...
d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)
Mà 37 = 2187 => x7 = 37 => x = 3
Vậy ....
e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)
Mà 38 = (-3)8 = 6561
=> x8 = 38 = (-3)8
=> x = {-3;3}
Vậy ...
PD
28 tháng 9 2018
nhé
a)(2x-1)6=(2x-1)8
=> (2x-1)8-(2x-1)6=0
=> (2x-1)6.((2x-1)2-1)=0
+)th1(2x-1)6=0
+)th2((2x-1)2-1)=0
TH
28 tháng 9 2018
a) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)\in\left\{\pm1;0\right\}\)
TH1 : \(2x-1=0\) TH2 : \(2x-1=-1\) TH3 : \(2x-1=1\)
\(2x=1\) \(2x=0\) \(2x=2\)
\(x=\frac{1}{2}\) \(x=0\) \(x=1\)
Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)
b) Tương tự
8 tháng 8 2017
1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)
8 tháng 8 2017
\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)
\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)
\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)
\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)
\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)
\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)
3) Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)
\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)
\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)
\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)
\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)
\(B< A\)