giúp mình với ạ
a) 2x-1.3y+1=12x+y b) 3x = 9y-1 và 8y = 2x+8
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\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
Vây \(S=\left\{x|x< \dfrac{15}{7}\right\}\)
lớp 8 chx hc kí hiệu đó anh ạ
a: =>2x-3x^2-x<15-3x^2-6x
=>x<-6x+15
=>7x<15
=>x<15/7
b: =>4x^2-24x+36-4x^2+4x-1>=12x
=>-20x+35>=12x
=>-32x>=-35
=>x<=35/32
a) Ta có: \(2^{x-1}\cdot3^{y+1}=12^{x+y}\)
\(\Leftrightarrow2^{x-1}\cdot3^{y+1}=4^{x+y}\cdot3^{x+y}\)
\(\Leftrightarrow2^{x-1}\cdot3^{y+1}=2^{2x+2y}\cdot3^{x+y}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=2x+2y\\y+1=x+y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-1=2\cdot1+2y\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2+2y=0\\x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=-2\\x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
b đâu bạn