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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)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
23 tháng 9 2016
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
19 tháng 6 2022
Câu 2:
\(2^{24}=8^8< 9^8=3^{16}\)
Câu 2:
c: =>x+5=-4
=>x=-9
d: =>2x-3=3 hoặc 2x-3=-3
=>2x=6 hoặc 2x=0
=>x=0 hoặc x=3
3 tháng 8 2018
a) 27x : 3x = 9
(27 : 3)x = 9
9x = 91
x = 1
b) 25 : 5x =5
5x = 25 : 5
5x = 51
x = 1
c) 2 : (x + 2)2 = \(\dfrac{1}{18}\)
(x + 2)2 = 2 : \(\dfrac{1}{18}\)
(x + 2)2 = 36
\(\Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
d) (5x - 1)2 = \(\dfrac{36}{49}\)
(5x - 1)2 = \(\left(\dfrac{6}{7}\right)^2\)
Bạn làm tiếp nha, mình có việc bận :v
25 tháng 2 2022
Bài 1:
a: \(\left(2x-1\right)^4=16\)
=>2x-1=2 hoặc 2x-1=-2
=>2x=3 hoặc 2x=-1
=>x=3/2 hoặc x=-1/2
b: \(\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}< =0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x+7=y=2\cdot3+7=13\end{matrix}\right.\)
c: \(10800=2^4\cdot3^3\cdot5^2\)
mà \(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)
nên \(\left\{{}\begin{matrix}x+2=4\\x+1=3\\x=2\end{matrix}\right.\Leftrightarrow x=2\)
NB
9 tháng 4 2020
\(A=\left(x-2\right)^2+2\)
Có: \(\left(x-2\right)^2\ge0với\forall x\\ \Rightarrow\left(x-2\right)^2+2\ge0\\ \Leftrightarrow A\ge0\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy....
\(B=\left(2x+1\right)^4-1\)
Có: \(\left(2x+1\right)^4\ge0với\forall x\\ \Rightarrow\left(2x+1\right)^4-1\ge-1\\ \Leftrightarrow B\ge-1\)
Dấu "=" xảy ra khi \(\left(2x+1\right)^4=0\Leftrightarrow x=-\frac{1}{2}\)
VẬy...
\(C=\left(x^2-16\right)^2+\left|y-3\right|-2\)
Có: \(\left(x^2-16\right)^2\ge0với\forall x\\ \left|y-3\right|\ge0với\forall x\\ \Rightarrow\left(x^2-16\right)^2+\left|y-3\right|-2\ge2\\ \Leftrightarrow C\ge2\)
Dấu "=" xảy ra khi \(\left(x^2-16\right)^2=0\Leftrightarrow x\in\left\{\pm16\right\}\); \(\left|y-3\right|=0\Leftrightarrow y=3\)
Vậy...
\(D=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
Có: \(\left(x+2\right)^2\ge0với\forall x\\ \left(y-\frac{1}{5}\right)^2\ge0với\forall x\\ \Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge-10\\ \Leftrightarrow D\ge-10\)
Dấu "=" xảy ra khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\);\(\left(y-\frac{1}{5}\right)^2=0\Leftrightarrow x=\frac{1}{5}\)
Vậy...
19 tháng 7 2017
a. 2x = 8 ; b. 5x = 25 ; c. 3x : 35 = 9 d. \(\dfrac{16}{2^x}=2\) ; e. 8x : 2x = 4 ; f. 2x . 3x = 36 ; g. \(\dfrac{\left(-3\right)^n}{81}=-27\)
2x = 23 5x = 52 3x : 35 = 32 \(\dfrac{2^4}{2^x}=1\) ( 23)x : 2x = 22 6x = 62 \(\dfrac{\left(-3\right)^n}{\left(-3\right)^4}=\left(-3\right)^3\)
x = 3 x = 3 3x = 32 . 35 \(2^{4-x}=2^1\) 23x : 2x = 22 x = 2 \(\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
3x = 37 \(\Rightarrow4-x=1\) 23x - x = 22 \(\left(-3\right)^n=\left(-7\right)^7\)
=>X = 7 x = 4 - 1 22x = 22 => n = 7
x = 3 2x = 2
x = 2 : 2
x = 1
30 tháng 7 2018
d)
\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)
\(=\left(\frac{6}{7^3}\right)^2\)
\(=\frac{6^2}{7^{3^2}}\)
\(=\frac{36}{7^6}\)
30 tháng 7 2018
\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)
\(=\left(\frac{6}{7^3}\right)^2\)
\(=\left(\frac{6^2}{7^{3^2}}\right)\)
\(=\frac{36}{7^6}\)
Code : Breacker
\(a,16^x:4^x=16\)
\(\left(4^x\right)^2:4^x=4^2\)
\(\Rightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,2^{-1}.2^x+4.2^x=72\)
\(\Rightarrow2^{x-1}+2^{x+2}=72\)
\(\Rightarrow2^{x-1}\left(1+2^3\right)=72\)
\(\Rightarrow2^{x-1}=72:9=8=2^3\)
\(\Rightarrow x=4\)
\(c,\left(2^x+1\right)^3=-64\)
\(\Rightarrow2^x+1=-4\)
\(\Rightarrow2^x=-5\)
a) 16^x : 4^x = 16
<=> ( 16 : 4 )^x = 16
<=> 4^x = 16
<=> 4^x = 4^2
=> x = 2
Vậy x =2