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a) (2x - 3)2 = 16
=> (2x - 3)2 = 42
=> \(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
=> \(\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy ...
b) 2.3x + 2 + 4. 3x + 1 = 10.36
=> 2.3x + 1. 3 + 4.3x + 1 = 10 . 36
=> 6.3x + 1 + 4.3x + 1 = 10.36
=> (6 + 4).3x + 1= 10.36
=> 10.3x + 1= 10.36
=> 3x + 1= 36
=> x + 1 = 6
=> x = 6 - 1
=> x = 5
\(a,\left(2x-3\right)^2=16\)
\(\Rightarrow\left(2x-3\right)^2=4^2\)
\(\Rightarrow2x-3=4\)
\(2x=4+3\)
\(2x=7\)
\(x=7:2\)
\(x=\frac{7}{2}\)
a) 23x+2 = 4x+5 = (22)x+5 = 22x+10
=> 3x + 2 = 2x + 10
=> 3x - 2x = 10 - 2
x = 8
b) 3-1.3x + 9.3x = 28
3x. ( 3-1 + 9) = 28
3x. 28/3 = 28
3x = 3 = 31
=> x = 1
\(2^{3x+2}=4^{x+5}\)
\(\Rightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Rightarrow3x+2=2\left(x+5\right)\)
\(\Rightarrow3x+2=2x+10\)
\(\Rightarrow3x-2x=10-2\Rightarrow x=8\)
\(3^{-1}.3^x+9.3^x=28\)
\(\Rightarrow\frac{1}{3}.3^x+9.3^x=28\)
\(\Rightarrow3^x.\left(9+\frac{1}{3}\right)=28\)
\(\Rightarrow3^x.\frac{28}{3}=28\)
\(\Rightarrow3^x=3\Rightarrow x=1\)
Chúc bạn học tốt.
1;Ta có\(5.3^x=5.3^4\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
2.Ta có \(9.5^x=6.5^6+3.5^6\)
\(\Rightarrow9.5^x=5^6.\left(6+3\right)\)
\(\Rightarrow9.5^x=9.5^6\)
\(\Rightarrow5^x=5^6\)\
\(\Rightarrow x=6\)
3, Ta có \(2.3^{x+2}+4.3^{x+1}=10.3^6\)
\(\Rightarrow3^{x+1}.\left(2.3+4\right)=10.3^6\)
\(\Rightarrow3^{x+1}.10=10.3^6\)
\(\Rightarrow3^{x+1}=3^6\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
a) 5.3x = 5.34
=> 3x=34
=> x=4
b) 9.5x=6.56+3.56
=> 9.5x = (6+3)56
=> 9.5x=9.56
=> 5x=56
=> x=6
c) 2.3x+2 + 4.3x+1 = 10.36
=> 2.3x+1.3 + 4.3x+1 = 10.36
=> 6.3x+1+4.3x+1=10.36
=> (6+4).3x+1=10.36
=> 10.3x+1=10.36
=> 3x+1=36
=> x+1=6
=> x=5
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
\(3^{-1}.3^x+9.3^x=28\\ \Leftrightarrow3^x\left(\dfrac{1}{3}+9\right)=28\\ \Leftrightarrow3^x.\dfrac{28}{3}=28\\ \Leftrightarrow3^x=3\\ \Leftrightarrow x=1\)
a,9<3x<39.3-2
\(\Leftrightarrow3^2< 3< 3^7\)
\(\Rightarrow S=\left\{3;4;5;6\right\}\)
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