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HE
Tìm x , biết:
a, ( 2x + 1)2 = 93
b, ( 2x + 1)3 = 1252
c, 252< 5x+3 < 254
d, x15= x
e, ( x - 5)4 = ( x-5)6
1
C
30 tháng 9 2016
A) \(\left(2x+1\right)^2=9^3\)
\(\Rightarrow\left(2x+1\right)^2=9^2\times9\)
\(\Rightarrow2x+1=81\)
\(\Rightarrow2x=81-1\)
\(\Rightarrow2x=80\)
\(\Leftrightarrow x=40\)
B) \(\left(2x+1\right)^3=125^2\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Leftrightarrow x=2\)
C) \(25^2< 5^{x+3}< 25^4\)
\(\Leftrightarrow5^4< 5^{x+3}< 5^6\)
\(\Leftrightarrow4< x+3< 6\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
D) \(x^{15}=x\)
Nếu \(x>1\)thì \(x^{15}>x\)
Vậy \(x=1\)
E) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Rightarrow\left(x-5\right)\times\left(1^4-1^6\right)=0\)
\(\Rightarrow x-5=0\)
\(\Leftrightarrow x=5\)
KÍCH MK NHA BẠN
29 tháng 8 2021
a) 154-(x-4)=2*25
154-x+4=50
x+4=104
x=100
b)2346:(25+x)=23
25+x=102
x=77
15 tháng 10 2017
a) \(2x^3-32x=0\)
\(2x\left(x^2-16\right)=0\)
\(2x\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow2x=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(x=0\) hoặc \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
b) \(\left(3x-2\right)^2-\left(x+5\right)^2=0\)
\(\left(3x-2-x-5\right)\left(3x-2+x+5\right)=0\)
\(\left(2x-7\right)\left(4x+3\right)=0\)
\(\orbr{\begin{cases}2x-7=0\\4x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
c) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
d) \(4x^2-25-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(4x^2-25\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x+5\right)\left(2x-5-x-7\right)=0\)
\(\left(2x+5\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
9 tháng 7 2018
a) Ta có: \(\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)
TH1: \(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
TH2: \(2x+\frac{3}{5}=\frac{-3}{5}\Rightarrow2x=\frac{-6}{5}\Rightarrow x=\frac{-3}{5}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
Mà \(\frac{-1}{27}=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\Leftrightarrow3x=\frac{1}{6}\Rightarrow x=\frac{1}{18}\)
c) \(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{2}{3}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{2}{3}x+\frac{5}{6}=0\)
\(\Rightarrow-\frac{37}{6}x=\frac{-1}{6}\Rightarrow x=\frac{1}{37}\)
d) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3-x-\frac{1}{5}=0\)
\(\Rightarrow-3x=\frac{47}{10}\Rightarrow x=\frac{-47}{30}\)
a, 2x - 20 = 35 : 32
=> 2x - 1 = 33
=> 2x - 1 = 27
=> 2x = 28
=> x = 14
b, 525.5x + 1 = 525
=> 525+x+1 = 525
=> 25 + x + 1 = 25
=> 26 + x = 25
=> x = -1
a. 2x - 20 = 35 : 32
2x - 1 = 33
2x = 27 + 1
2x = 28
x = 28 : 2
x = 14.