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a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
\(a,\left(x-3\right)^2=1\)
\(\left(x-3\right)^2=1^2=\left(-1\right)^2\)
\(\hept{\begin{cases}x-3=1\\x-3=-1\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=2\end{cases}}}\)
\(b,\left(2x-1\right)^3=27\)
\(\left(2x-1\right)^3=3^3\)
\(2x-1=3\)
\(2x=4\)
\(x=2\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
\(\hept{\begin{cases}x+\frac{1}{2}=\frac{1}{2}\\x+\frac{1}{2}=-\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(d,\left(3x-1\right)^3=0\)
\(3x-1=0\)
\(3x=1\)
\(x=\frac{1}{3}\)
a,(x-3)2=1
=>x2-9=1
=>x2 = 10
=> x=\(\sqrt{10}\)
b,(2x-1)3=27
=>8x3-1=27
=>8x3=28
=>x3=3,5
=>x=\(\sqrt{3,5}\)
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
a/ \(\left(\frac{1}{5}\right)^x=\left(\frac{1}{5^3}\right)^3=\left(\frac{1}{5}\right)^9\Rightarrow x=9\)
b/ \(\left(\frac{3}{5}\right)^x=\left(\frac{3^2}{5^2}\right)^3=\left(\frac{3}{5}\right)^6\Rightarrow x=6\)
c\(2^{3-2x}=\left(2^3\right)^3=2^9\Rightarrow3-2x=9\Rightarrow x=-3\)
d/ \(2^{3x+1}=32^2=\left(2^5\right)^2=2^{10}\Rightarrow3x+1=10\Rightarrow x=3\)
e/ \(3^{6-3x}=81^3=\left(3^4\right)^3=3^{12}\Rightarrow6-3x=12\Rightarrow x=-2\)
\(\left(\frac{1}{5}\right)^x=\left(\frac{1}{125}\right)^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left[\left(\frac{1}{5}\right)^3\right]^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left(\frac{1}{5}\right)^9\Leftrightarrow x=9\)
\(\left(\frac{3}{5}\right)^x=\left(\frac{9}{25}\right)^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left[\left(\frac{3}{5}\right)^2\right]^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^6\Leftrightarrow x=6\)
\(2^{3-2x}=8^3\Leftrightarrow2^{3-2x}=\left(2^3\right)^3\Leftrightarrow2^{3-2x}=2^9\Leftrightarrow3-2x=9\)
\(\Leftrightarrow2x=3-9\Leftrightarrow2x=-6\Leftrightarrow x=\left(-6\right):2\Leftrightarrow x=-3\)
Các phép còn lại làm tương tự bn nha !
a) \(\frac{27}{3x}=3\)
=> \(27=3.3x\)
=> \(27=9x\)
=> \(x=27:9\)
=> \(x=3\)
Vậy \(x=3.\)
b) \(\left(2x-1\right)^3=-125\)
=> \(\left(2x-1\right)^3=\left(-5\right)^3\)
=> \(2x-1=-5\)
=> \(2x=\left(-5\right)+1\)
=> \(2x=-4\)
=> \(x=\left(-4\right):2\)
=> \(x=-2\)
Vậy \(x=-2.\)
Chúc bạn học tốt!
a) đk x khác 0.Theo đề bài:
\(27=3.3x=9x\Rightarrow x=3\)
b) \(\left(2x-1\right)^3=-125=-5^3\Rightarrow2x-1=-5\Rightarrow x=-2\)
c) Sửa đề \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\) và x thuộc Z(cho khó thêm:v nhưng mà ko chắc đâu;v. Sai thì bảo mình làm lại theo đề cũ của bạn)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^{x+2}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x-1\right)^{x+2}=1\left(1\right)\end{matrix}\right.\). Dễ thấy x = 0 thỏa mãn (1).
Với x < 0 nhận xét rằng khi đó lũy thừa số số mũ chẵn luôn lớn hơn lũy thừa có số mũ lẽ. Do đó \(\left(x-1\right)< -1;x+2< 2\Rightarrow\left(x-1\right)^{x+2}< \left(-1\right)^2=1\) (không cần xét mũ lẻ do cái nhận xét trên)
Với x > 0 thì:
+) Với x lẻ đặt x = 2k + 1 (k>=0)=> x + 2 = 2k + 3 và x - 1 = 2k
+) Với x chẵn dặt x = 2k (k>=0) => x + 2 = 2k + 2; x - 1 =2k -1
Khi đó ta sẽ suy ra nhận xét sau với x > 0 thì (x-1)x+2 với x lẻ > (x-1)x+2 với x chẵn.
Nên \(\left(x-1\right)^{x+2}>\left(2k-1\right)^{2k+2}\ge\left(2.0-1\right)^{2.0+2}=1\) nên nó vô nghiệm
Vậy x = 0 hoặc x = 1