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1)
a) \(0,25^x\cdot12^x=243\)
\(\Leftrightarrow\left(0,25\cdot12\right)^x=3^5\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(38^y:19^y=512\)
\(\Leftrightarrow2y\cdot y=512\)
\(\Leftrightarrow2y^2=512\)
\(\Leftrightarrow y^2=256\)
\(\Leftrightarrow\left[{}\begin{matrix}y=16\\y=-16\end{matrix}\right.\)
Vậy \(y_1=-16;y_2=16\)
2)
a) \(3^x+3^{x+2}=2430\)
\(\Leftrightarrow\left(1+3^2\right)\cdot3^x=2430\)
\(\Leftrightarrow\left(1+9\right)\cdot3^x=2430\)
\(\Leftrightarrow10\cdot3^x=2430\)
\(\Leftrightarrow3^x=243\)
\(\Leftrightarrow3^x=3^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(2^{x+3}-2^x=224\)
\(\Leftrightarrow\left(2^3-1\right)\cdot2^x=224\)
\(\Leftrightarrow\left(8-1\right)\cdot2^x=224\)
\(\Leftrightarrow7\cdot2^x=224\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
3)
a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\pm\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}+\dfrac{1}{4}\\x=-\dfrac{2}{3}+\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{11}{12};x_2=-\dfrac{5}{12}\)
b) \(\left(x+0,7\right)^3=-27\)
\(\Leftrightarrow\left(x+\dfrac{3}{10}\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x+\dfrac{3}{10}=-3\)
\(\Leftrightarrow x=-3-\dfrac{3}{10}\)
\(\Leftrightarrow x=-\dfrac{37}{10}\)
Vậy \(x=-\dfrac{37}{10}\)
4)
a) \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{2}{5}-3x=\pm\dfrac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{15};x_2=\dfrac{1}{3}\)
b) \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(\Leftrightarrow2x-1=1\)
\(\Leftrightarrow2x=1+1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
1. a) \(0,25^x.12^x=243\)
\(\Rightarrow\left(0,25.12\right)^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
b) \(38^y:19^y=512\)
\(\Rightarrow\left(38:19\right)^y=512\)
\(\Rightarrow2^y=2^9\)
\(\Rightarrow y=9\)
Vậy \(y=9.\)
2) a) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x\left(1+9\right)=2430\)
\(\Rightarrow3^x=243=3^5\)
\(\Rightarrow x=5\)
Vậy x=5.
b) \(2^{x+3}-2^x=224\)
\(\Rightarrow2^x\left(8-1\right)=224\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
Vậy x=5.
Bài 3: dễ tự làm.
a) 2x+1.3y=123
<=>2x+1.3y=(22)3.33
<=> 2x+1=26 và 3y=33
<=>x+1=6 và y=3
<=>x=5 và y=3
b) 10x : 5y=20y
<=>10x=20y.5y=100y=(102)y
<=>x=2y (Nhiều số lắm chèn)
c) 2x=4y-1
<=>2x=2y-2
<=>x=y-2
Mặt khác: 27y=3x+8
<=> 33y=3x+8
<=>3y=x+8
<=>3y=(y-2)+8
<=>2y=6
<=>y=3
=>x=y-2=3-2=1
a)0,25x.12x=243
(0,25x12)x=243
3x=243
3x=35
=>x=5
tìm các số tự nhiên x, y biết: (x+3).(y-5)=38
giúp tôi trả lời câu này với!