Tìm x:
\(x^5\)=\(5^x\)
HELP ME, PLEASE!
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Ta có: \(|-x-5|=|1-5|\)
<=> \(|-x-5|=4\)
<=>\(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)
<=>\(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)
Vậy x=1 hoặc x=9
\(5^{x+2}-5^x=3.10^3\)
\(\Rightarrow5^x\left(5^2-1\right)=3000\)
\(\Rightarrow5^x.24=3000\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
\(5^{x+2}-5^x=3\cdot10^3\)
\(\Leftrightarrow5^x\cdot24=3000\)
\(\Leftrightarrow x=3\)
a: \(=\dfrac{7}{5}\cdot\dfrac{15}{49}-\dfrac{12+10}{15}:\dfrac{11}{5}\)
\(=\dfrac{3}{7}-\dfrac{22}{15}\cdot\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=\dfrac{-5}{21}\)
b: =>2,8x-32=-60
=>2,8x=-28
hay x=-10
\(\dfrac{5}{7}< \dfrac{x}{5}< 1\)
\(\dfrac{5\times5}{7\times5}< \dfrac{7\times x}{7\times5}< \dfrac{35}{35}\)
\(\dfrac{25}{35}< \dfrac{7\times x}{35}< \dfrac{35}{35}\)
\(25< 7\times x< 35\)
\(\dfrac{25}{7}< x< \dfrac{35}{7}\)
\(\dfrac{25}{7}< x< 5\)
\(3< x< 5\) (x is the natural number)
x=4
/x+5/+/x-1/=12:/y+5/+3
suy ra x+5+x-1=12:/y+5/+3 hoặc x+5+x-1=12:-<y+5>+3
\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)
\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)
Mình làm câu đầu tượng trưng thui nhé, 2 câu sau tương tự vậy !!!!!!
a) pt <=> \(x^2-2xy+2y^2-2x-2y+5=0\)
<=> \(\left(x-y-1\right)^2+y^2-4y+4=0\)
<=> \(\left(x-y-1\right)^2+\left(y-2\right)^2=0\) (1)
TA LUÔN CÓ: \(\left(x-y-1\right)^2;\left(y-2\right)^2\ge0\forall x;y\)
=> \(\left(x-y-1\right)^2+\left(y-2\right)^2\ge0\) (2)
TỪ (1) VÀ (2) => DẤU "=" SẼ PHẢI XẢY RA <=> \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)
VẬY \(\left(x;y\right)=\left(3;2\right)\)
\(x^5=5^x\)
\(\Rightarrow x=5\)
x=5 nha
hok tốt !!!