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a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)
⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\)
⇔ 2x + 2 - 5 - 2x = 12 -16x
⇔ 16x = 15
⇔ x = 15/16
b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)
⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)
⇔ 8 - 6x - 4 + x = 5x + 10
⇔ 10x = -6
⇔ x = -6/10
Câu 1:
x + 1/4 - 5 + 2x/8 = 3 - 4x/2
<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8
<=> 2x + 2 - 5 - 2x = 12 - 16x
<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16
Câu 2:
4 - 3x/5 - 4 - x/10 = x + 2/2
<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10
<=> 8 - 6x - 4 + x = 5x + 10
<=> 4 - 5x = 5x + 10
<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5
\(a,\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(c,\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(e,\left(x-4\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`a,(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
`b,(x -2)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`c,(x +3)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
`d,(x + 1/2)(4x + 4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
`e,(x -4)(5x -10)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
`f,(2x -1)(3x +6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`g,(2,3x -6,9)(0,1x -2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)
\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)
d, ĐKXĐ:\(x\ne-2,x\ne3\)
\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)
1) * Xét \(x\ge-8\) thì \(x+8\ge0\)nên \(|x+8|=x+8\)
Đặt PT là A
A trở thành: x+8=4x-10
\(\Leftrightarrow x-4x=-10-8\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=\frac{-18}{-3}=6\)( thỏa ĐK vì x>-8)
* Xét \(x< -8\)thì\(x+8< 0\)nên \(|x+8|=-\left(x+8\right)=-x-8\)
A trở thành: \(-x-8=4x-10\)
\(\Leftrightarrow-x-4x=-10+8\)
\(\Leftrightarrow-5x=-2\)
\(\Leftrightarrow x=\frac{-5}{-2}=\frac{5}{2}\)(không thỏa Đk vì 5/2>-8)
Vậy tập nghiệm của PT đã cho là: S={6}
2) * Xét \(x\ge9\)thì\(x-9\ge0\)nên \(|x-9|=x-9\)
ĐẶT PT ĐỀ CHO LÀ B
B trở thành:\(x-9=2x+13\)
\(\Leftrightarrow x-2x=13+9\)
\(\Leftrightarrow-x=22\)
\(\Leftrightarrow x=-22\)(không thòa Đk do x<9)
*Xét \(x< 9\)thì\(x-9< 0\)nên \(|x-9|=-\left(x-9\right)=9-x\)
B trở thành:9-x=2x+13
\(\Leftrightarrow-x-2x=13-9\)
\(\Leftrightarrow-3x=4\)
\(\Leftrightarrow x=\frac{4}{-3}=\frac{-4}{3}\)(thỏa Đk vì x<9)
Vậy tập nghiệm của PT đã cho là: S={-4/3}
giúp bạn được nhiêu đó tk mk nha
1) Ta có: \(4x+8=3x-1\)
\(\Leftrightarrow4x-3x=-1-8\)
\(\Leftrightarrow x=-9\)
2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)
\(\Leftrightarrow10-5x-15-3x+3>0\)
\(\Leftrightarrow-8x>2\)
hay \(x< \dfrac{-1}{4}\)
a)\(3x-1-5\left(x+2\right)=x-4\)
\(\Leftrightarrow3x-1-5x-10=x-4\)
\(\Leftrightarrow3x-5x-x=-4+1+10\)
\(\Leftrightarrow-3x=7\)
\(\Leftrightarrow x=-\frac{7}{3}\)
a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)
=>x(2x+10-x+2)=0
=>x(x+12)=0
=>x=0 hoặc x=-12
b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
=>(2x-5)(3x+12)=0
=>x=5/2 hoặc x=-4
c: \(\Leftrightarrow\left(2x\right)^2-\left(x+3\right)^2=0\)
=>(x-3)(3x+3)=0
=>x=3 hoặc x=-1
d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
=>(x+2)(-5x+3)=0
=>x=-2 hoặc x=3/5
\(a,\left(x-2\right)x=2x\left(x+5\right)\)
\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)
\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)
Đề bị sao rồi bạn