x\(^2\)-\(\frac{3}{5}\)x=0
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Theo đề: \(2x+y=0\Leftrightarrow y=-2x\) \(\left(1\right)\)
Ta có:
\(\dfrac{3-x}{y-4}=\dfrac{2}{5}\)
\(\Leftrightarrow5\left(3-x\right)=2\left(y-4\right)\)
\(\Leftrightarrow15-5x=2y-8\)
\(\Leftrightarrow15+8=2y+5x\)
\(\Leftrightarrow5x+2y=23\) \(\left(2\right)\)
Thế (1) vào (2), suy ra:
\(5x+2.\left(-2x\right)=23\)
\(\Leftrightarrow5x-4x=23\)
\(\Leftrightarrow x=23\)
\(\Rightarrow y=-2.23=-46\)
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)
\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(x-1\right)\left(x+5\right)-4\left(x-1\right)=0\)
=>(x-1)(x+3)=0
=>x=1 hoặc x=-3
\(\Leftrightarrow2\left(x-1\right)\left(x+5\right)-4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
\(\frac{1}{2}x+\frac{3}{5}.\left(x-2\right)=3\)
\(\frac{1}{2}.x+\frac{3}{5}.x-\frac{6}{5}=3\)
\(\frac{11}{10}x-\frac{6}{5}=3\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{42}{11}\)
các bạn ơi mình đang cần gấp . Mình chỉ còn 20 phút thui . HUHU
Bài làm :
1) \(\frac{x+11}{4}=\frac{2x+4}{5}\)
\(\Leftrightarrow\left(x+11\right).5=4.\left(2x+4\right)\)
\(\Leftrightarrow5x+55=8x+16\)
\(\Leftrightarrow5x-8x=16-55\)
\(\Leftrightarrow-3x=-39\)
\(\Leftrightarrow x=\frac{-39}{-3}=\frac{39}{3}=13\)
2)\(\frac{x+4}{x+10}=\frac{3}{5}\)
\(\Leftrightarrow\left(x+4\right).5=\left(x+10\right).3\)
\(\Leftrightarrow5x+20=3x+30\)
\(\Leftrightarrow5x-3x=30-20\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=\frac{10}{2}=5\)
3)\(\frac{x+8}{x+14}=\frac{2}{3}\)
\(\Leftrightarrow\left(x+8\right).3=\left(x+14\right).2\)
\(\Leftrightarrow3x+24=2x+28\)
\(\Leftrightarrow3x-2x=28-24\)
\(\Leftrightarrow x=4\)
\(x^2-\frac{3}{5}x=0\)
=> \(xx-\frac{3}{5}x=0\)
=>\(x\left(x-\frac{3}{5}\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{3}{5}=0\\x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0+\frac{3}{5}\\x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=0\end{cases}}\)
Vậy x = 0 hoặc x = \(\frac{3}{5}\)
\(x^2-\frac{3}{5}x=0\)
\(\Leftrightarrow x\left(x-\frac{3}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-\frac{3}{5}=0\Leftrightarrow x=\frac{3}{5}\end{cases}}\)
Vậy x = 0 hoặc x = \(\frac{3}{5}\)