Tìm \(x\), biết :
a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5},\left(x\ne0\right)\)
b) \(x^{10}=25x^8\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DM
1
1 tháng 7 2017
Giải:
a) Ta có:
\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\left(x^7\ne0\right)\Leftrightarrow x=1\)
Vậy \(x=1\)
b) Ta có:
\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow\) \(\left[\begin{array}{}x^8=0\\x^2-25=0\end{array}\right.\)
\(\Leftrightarrow\) \(\left[\begin{array}{}x=0\\x=5\\x=-5\end{array}\right.\) Vậy...
TH
2
12 tháng 7 2016
a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\)
\(\Rightarrow x^8=x^7\)
\(\Rightarrow x^8-x^7=0\)
\(\Rightarrow x^7.\left(x-1\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(x^7\ne0\) )
Vậy \(x=1\)
b ) \(x^{10}=25x^8\)
\(\Rightarrow x^{10}-25x^8=0\)
\(\Rightarrow x^8.\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8=0\) hoặc \(x^2-25=0\)
Do đó \(x=0\) hoặc \(x=5\) hoặc \(x=-5\)
Vậy \(x\in\left\{0;5;-5\right\}\)
O
18 tháng 4 2022
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
VN
22 tháng 7 2018
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
VN
22 tháng 7 2018
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
14 tháng 8 2023
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
14 tháng 8 2023
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
DT
Dang Tung
CTVHS
10 tháng 10 2023
a) (-5/9)^10 : x = (-5/9)^8
=> x = (-5/9)^10 : (-5/9)^8
=> x = (-5/9)^10-8 = (-5/9)^2
=> x = 25/81
b ) x : (-5/9)^8 = (-9/5)^8
=> x = (-9/5)^8 . (-5/9)^8
=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )
=> x = 1
C) x^3 = -8 =(-2)^3
=> x = -2
KV
10 tháng 10 2023
a) (-5/9)¹⁰ : x = (-5/9)⁸
x = (-5/9)¹⁰ : (-5/9)⁸
x = (-5/9)²
x = 25/81
b) x : (-5/9)⁸ = (-9/5)⁸
x = (-9/5)⁸ . (-5/9)⁸
x = [-9/5 . (-5/9)]⁸
x = 1⁸
x = 1
c) x³ = -8
x³ = (-2)³
x = -2
T
1 tháng 9 2023
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
T
1 tháng 9 2023
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
7 tháng 10 2021
\(a,=5^3:5^2=5\\ b,=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\\ c,=1728-512=1216\\ d,=x^{10}:x^8=x^2\\ e,=\left(-x\right)^{5-3}=\left(-x\right)^2=x^2\\ f,=\left(-y\right)^{5-4}=-y\)
a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\\ x^8=x^7\\ \Rightarrow x=1;x=-1\)
b)\(x^{10}=25.x^8\\ x^2=25\\ \Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
a) \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)
\(\Rightarrow x^8=x^7\)
\(\Rightarrow x^8-x^7=0\)
\(\Rightarrow x^7.x-x^7=0\)
\(\Rightarrow x^7\left(x-1\right)=0\)
\(\Rightarrow x-1=0\) (vì x^7 \(\ne\)0)
\(\Rightarrow\) x=1
b) x^10=25x^8
\(\Rightarrow x^8.x^2-25x^8=0\)
\(\Rightarrow x^8\left(x^2-25\right)=0\)
\(\Rightarrow x^8=0\) hoặc \(x^2-25=0\)
1) x^8=0
\(\Rightarrow\) x=0(1)
2) x^2 -25=0
x^2=0+25
x^2=25
x^2=5^2 hay x^2=(-5)^2
Suy ra x=5 hoặc x=-5 (2)
Từ (1) và (2)\(\Rightarrow\)x\(\in\left\{0;5;-5\right\}\)
EM KO CHÉP ĐÁP ÁN NHÉ