2^x=144-2^x+3
x.(x^2)^5=x^5
(x-5)^2+ly^2-4l=0
(x-0,2)^10+(y+3,1)^20=0
3x=4y=6z và x^2+y^2=25
3x=2y ;7y=5z và x-y+z=32
x+1/x+5=x-1/x-3
x-1/2=y-3/4=z-5/6 và 2x+5y-3z=32
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a, \(9x=4y\Rightarrow\frac{x}{4}=\frac{y}{9}=\frac{y-x}{9-4}=\frac{-25}{5}=-5\)
\(\Rightarrow\hept{\begin{cases}x=-5\times4=-20\\y=-5\times9=-45\end{cases}}\)
b,\(\frac{x}{2}=\frac{y}{5}=\frac{3x}{6}=\frac{2y}{10}=\frac{3x-2y}{6-10}=\frac{20}{-4}=-5\)
\(\Rightarrow\hept{\begin{cases}x=-5\times2=-10\\y=-5\times5=-25\end{cases}}\)
c,\(\frac{x}{3}=\frac{y}{5}\Rightarrow\frac{x^2}{9}=\frac{y^2}{25}=\frac{x^2-y^2}{9-25}=\frac{-64}{-16}=4\)
\(\Rightarrow\hept{\begin{cases}x^2=9\times4=36\\y^2=25\times4=100\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm6\\y=\pm10\end{cases}}\)
Ta thấy \(\frac{x}{3}=\frac{y}{5}\)nên x,y cùng dấu
Vậy ....................................................
d, \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\);\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{15}=\frac{z}{18}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{18}\)từ đó bạn tự giải nha
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
\(\frac{2}{3x}=\frac{3}{4y}=\frac{5}{6z}\Rightarrow\frac{2}{30.3x}=\frac{3}{30.4y}=\frac{5}{30.6z}\Leftrightarrow\frac{1}{45x}=\frac{1}{40y}=\frac{1}{36z}\Rightarrow45x=40y=36z\)
\(\Rightarrow x=\frac{9}{8}y;x=\frac{5}{4}z\Rightarrow x^2+y^2+z^2=x^2+\frac{64}{81}x^2+\frac{16}{25}x^2\)
\(=x^2\left(1+\frac{2896}{2025}\right)=724\text{ :)) đến đây thôi :))}\)