x - 8 (15 + 5x) = 0
giup minh nhe minh dang can gap ...
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72-3x=5x+8
=>-3x-5x=8-72
=>(-3-5).x=-64
=>-8x=-64
=>x=(-64):(-8)
=>x=8
\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)
\(\left(4-8x\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)
\(2\left(x-4\right)-3x\left(x-4\right)=0\)
\(\left(2-3x\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
\(B=...\)
\(=\left(4+12+20+...+92\right)+\left(8+16+24+...+96\right)-\left(8+16+...+96\right)\)
\(=4+12+20+...+92\)
\(=\left(4+92\right)\times\left[\frac{\left(92-4\right)}{8}+1\right]=96\times12=1152\)
\(\left(x+2\right)2+\left(x-3\right)=0\)
\(2x+4+x-3=0\)
\(\left(2x+x\right)+\left(4-3\right)=0\)
\(3x+1=0\)
Nếu em chỉ mới học đến lớp 5 thì đến đây là không giải được ! Nếu học rồi thì làm tiếp !
\(3x=0-1\)
\(3x=-1\)
\(x=-\frac{1}{3}\)
a, Có \(\dfrac{3x-2y}{7}=\dfrac{4x+3y}{5}\)
=> 5(3x-2y)=7(4x+3y)
=> 15x-10y=28x+21y
=> 15x-28x=21y+10y
=> -13x=31y
=> \(\dfrac{x}{y}=\dfrac{31}{-13}=\dfrac{-31}{13}\)
b,\(\dfrac{5x-2y}{3x+4y}=\dfrac{-3}{4}\)
=> 4(5x-2y)=-3(3x+4y)
=> 20x-8y= -9x-12y
=> 20x+9x=-12y+8y
=> 29x=-4y
=> \(\dfrac{x}{y}=\dfrac{-4}{29}\)
x-120-40x =0
-120-39x=0
-39x=120
x=120:-39
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