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Bài 1:
b) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-\frac{1}{2}\\x=0+\frac{3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{4}\right\}.\)
c) \(\left(2x-5\right)^4=81\)
\(\Rightarrow2x-5=\pm3\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3+5=8\\2x=\left(-3\right)+5=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8:2\\x=2:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{4;1\right\}.\)
d) \(3^{x+1}+3^{x+3}=810\)
\(\Rightarrow3^x.3^1+3^x.3^3=810\)
\(\Rightarrow3^x.\left(3^1+3^3\right)=810\)
\(\Rightarrow3^x.30=810\)
\(\Rightarrow3^x=810:30\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
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)\(5^x.\left(5^3\right)^2=625\)
\(5^x.5^6=5^4\)
\(5^x=5^{-2}\)
\(x=-2\)
b)\(27< 81^3:3^x< 243\)
\(3^3< \left(3^4\right)^3:3^x< 3^5\)
\(3^3< 3^{12}:3^x< 3^5\)
\(3^{12}:3^x=3^4\)
\(3^x=3^3\)
\(x=3\)
c)\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(5x+1=\frac{6}{7}\)
\(5x=\frac{-1}{7}\)
\(x=\frac{-1}{35}\)
d)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{6}{9}=\frac{2}{3}\)
\(5^x.\left(5^3\right)^2=625\)
\(\Rightarrow5^x.5^6=5^4\)
\(\Rightarrow5^{x+6}=5^4\Rightarrow x+6=4\Rightarrow x=-2\)
Đề sai rồi bạn : Phải là :
\(5^x:\left(5^3\right)^2=625\)
\(\Rightarrow5^x:5^6=5^4\)
\(\Rightarrow5^{x-6}=5^4\)
\(\Rightarrow x-6=4\Rightarrow x=10\)
Nhứng nếu đề đúng thì bạn có thể lấy KQ trên
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 !
\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)
\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)
\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)
\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)
\(\Leftrightarrow\text{ }x=\frac{53}{45}\)
\(\text{Vậy }x=\frac{53}{45}\)
\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)
\(\text{TH 1: Dấu chia}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)
\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)
\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)
\(\text{Vậy }\text{}x=\frac{-1}{5}\)
\(\text{TH 2:Dấu cộng}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{-3}{25}\)
\(\Leftrightarrow x=\frac{-3}{25}:2\)
\(\Leftrightarrow x=\frac{-3}{50}\)
\(\text{Vậy }x=\frac{-3}{50}\)
\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)
\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)
\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)
\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{5}{4}\)
\(\text{Vậy }x=\frac{5}{4}\)
\(a)\)\(\left(\frac{3}{5}\right)^{2x+1}=\frac{81}{625}\)
\(\Leftrightarrow\)\(\left(\frac{3}{5}\right)^{2x+1}=\left(\frac{3}{5}\right)^4\)
\(\Leftrightarrow\)\(2x+1=4\)
\(\Leftrightarrow\)\(x=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
\(b)\)\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)^3=\frac{32}{243}\)
\(\Leftrightarrow\)\(\left(\frac{2}{3}\right)^{x+3}=\left(\frac{2}{3}\right)^5\)
\(\Leftrightarrow\)\(x+3=5\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(c)\)\(\left(2x-1\right)^2=\left(2x-1\right)^3\)
\(\Leftrightarrow\)\(\left(2x-1\right)^3-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-1-1\right)=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(2x-1\right)^2=0\\2x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=1\)
Chúc bạn học tốt ~