tìm x biết x2/ 6= 24/25
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\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!
20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
\(25\cdot x^2=25\\ x^2=25\div25\\ x^2=1\\ x^2=1^2\\ x=1\\ 8^{x-3}=1\\ 8^{x-3}=8^0\\ x-3=0\\ x=0+3\\ x=3\)
`@` `\text {Ans}`
`\downarrow`
\(3^{x+6}=37\)
`=>`\(3^x\cdot3^6=37\)
`=>`\(3^x=37\div3^6\)
`=>` \(3^x=\dfrac{37}{729}\)
Bạn xem lại đề.
\(25x^2=25\)
`=>`\(x^2=25\div25\)
`=>`\(x^2=1\)
`=> x=1`
\(8^{x-3}=1\)
`=>`\(8^x\div8^3=1\)
`=>`\(8^x=8^3\)
`=> x=3`
\(Ư\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\\ \Rightarrow x=5\left(B\right)\\ B\left(8\right)=\left\{0;8;16;24;32;...\right\}\\ \Rightarrow x=24\left(B\right)\)
\(1+5^2+5^4+...+5^{2x}\left(1\right)=\dfrac{25^6-1}{24}\)
Đặt \(\left(1\right)=A\)
\(\Rightarrow A=1+5^2+...+5^{2x}\)
\(\Rightarrow5^2A=5^2+5^4+...+5^{2x+2}\)
\(\Rightarrow25A=5^2+5^4+...+5^{2x+2}\)
\(\Rightarrow25A-A=5^2+5^4+...+5^{2x+2}-1-5^2-...-5^{2x}\)
\(\Rightarrow24A=5^{2x+2}-1\)
\(\Rightarrow A=\dfrac{5^{2x+2}-1}{24}\)
Mà: \(A=\dfrac{25^6-1}{24}\)
\(\Rightarrow\dfrac{5^{2x+2}-1}{24}=\dfrac{\left(5^2\right)^6-1}{24}\)
\(\Rightarrow5^{2x+2}-1=5^{12}-1\)
\(\Rightarrow5^{2x+2}=5^{12}\)
\(\Rightarrow2x+2=12\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=\dfrac{10}{2}\)
\(\Rightarrow x=5\)