Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: =>(x-2)(3x+1)-(x-2)(x+2)=0
=>(x-2)(3x+1-x-2)=0
=>(x-2)(2x-1)=0
=>x=1/2 hoặc x=2
b: =>3(x-1)+4(x+1)=6(x-1)
=>3x-3+4x+4=6x-6
=>7x+1=6x-6
=>x=-7
c: =>x(x-3)-(x+2)(x+3)+16=0
=>x^2-3x-x^2-5x-6+16=0
=>10-8x=0
=>x=5/4
a: =>3(3x-7)+2(x+1)=-96
=>9x-21+2x+2=-96
=>11x-19=-96
=>11x=-96+19=-75
=>x=-75/11
b: \(x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x+1)=3(2x+1)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
a)
\(\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\)
\(< =>9x-21+2x+2=-96\)
\(< =>9x+2x=-96+21-2\\ < =>11x=-77\\ < =>x=-7\)
b)
\(\dfrac{x-x+1}{3}=\dfrac{2x+1}{5}\\ < =>5=6x+3\\ < =>6x=5-3\\ < =>6x=2\\ < =>x=\dfrac{1}{3}\)
c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
`b,(x+5)(2x-3)=0`
`<=>` $\left[ \begin{array}{l}x+5=0\\2x-3=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-5\end{array} \right.$
Vậy `S={-5,3/2}`
1)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(x\times\left(x+2\right)-3\times\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\times\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
c) \(\frac{x-6}{x+1}=\frac{x^2}{x-1}\)
nhân chéo lên, ngại chết đc
heoheo lần sau bạn đánh = kí hiệu đi :(((
a/ \(\dfrac{x}{3}+\dfrac{2x-1}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow2x+2x-1=3\)
<=> 4x = 4 <=> x = 1
Vậy x = 1
b/ \(\dfrac{3x+1}{2}+\dfrac{x-1}{3}=\dfrac{x-9}{6}\)
\(\Leftrightarrow3\left(3x+1\right)+2\left(x-1\right)=x-9\)
\(\Leftrightarrow9x+3+2x-2=x-9\)
\(\Leftrightarrow10x=-10\Leftrightarrow x=-1\)
Vậy pt có nghiệm x = -1
c/ \(\dfrac{x-1}{x-2}=\dfrac{x+3}{x+2}\) ĐKXĐ: \(x\ne\pm2\)
<=> \(\left(x-1\right)\left(x+2\right)=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+2x-x-2=x^2-2x+3x-6\)
\(\Leftrightarrow0x=-4\left(voly\right)\)
Vậy pt vô nghiệm
d/ \(\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-3\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)
pt <=> \(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}=\dfrac{2\left(3x+1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}\)
=> (3x-1)(x+3) + (x-3)(3x+1) = 2(3x+1)(x+3)
\(\Leftrightarrow3x^2+8x-3+3x^2-8x-3=6x^2+20x+6\)
\(\Leftrightarrow-20x=12\Leftrightarrow x=-\dfrac{3}{5}\left(tm\right)\)
Vậy pt có nghiệm x=....
e/ như ý d
Mơn bn nhe ^^ tại mjk chưa bt ạk