5-2/4x^2-2x-1/8x-16=x-1/2x(x-2)-7/8x
5-2/4x^2-2x-1/8x-16=x-1/2x(x-2)-7/8x
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\)
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )
\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)
\(\Rightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow-3x+2x+x=10+4-14\)
\(\Leftrightarrow0=0\)
Vậy pt đã cho có nghiệm đúng với mọi x
giải phương trình: x-1/2x^2-4x - 7/8x = 5-x/4x^2-8x - 1/8x-16
Trả lời:
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)
\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)
\(\Leftrightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow10-3x=10-3x\)
\(\Leftrightarrow-3x+3x=10-10\)
\(\Leftrightarrow0x=0\)( luôn thỏa mãn )
Vậy S = R với \(x\ne0;x\ne2\)
\(\dfrac{7}{8x}+\dfrac{5-x}{4x^2-8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
\(\dfrac{7}{8x}+\dfrac{5-x}{4x^2-8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
ĐKXĐ: x ≠ 0; x ≠ 2
\(< =>\dfrac{14x-28+20-4x}{16x\left(x-2\right)}=\dfrac{8x-8+2x}{16x\left(x-2\right)}\)
Suy ra: 14x - 28 + 20 - 4x = 8x - 8 + 2x
<=> 14x - 8x - 2x - 4x = 28 - 20 - 8
<=> 0x = 0
Vậy: S = { x | x ≠ 0;2 }
5-x/4x^2+8x+7/8=x-1/2x(x-2)+1/8x-16
Giải phương trình
a) \(\frac{4}{20-6x-2x^2}\)+ \(\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
b)\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2-10x}+10=\frac{x+25}{2x^2-50}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
c) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x.\left(x-2\right)}+\frac{1}{8x-16}\)
a) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x^2-4x}+\frac{1}{8x-16}\)
b) 3x.|x+1|−2x|x+2|=12
<=> \(\frac{7}{8x}+\frac{5-x}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\)(DK: x khác 0 và 2)
<=>\(\frac{7x\left(x-2\right)}{8x\left(x-2\right)}+\frac{10-2x}{8x\left(x-2\right)}=\frac{4x-4}{8x\left(x-2\right)}=\frac{x}{8x\left(x-2\right)}\)
<=>\(7x^2-14x+10-2x=4x-4+x\)
<=>\(7x^2-14x-2x-4x-x=-4-10\)
<=>\(7x^2-21x+14=0\)
<=>\(7\left(x^2-3x+2\right)=0\)
<=>\(x^2-3x+2=0\)
<=>\(x^2-x-2x+2=0\)
<=>\(x\left(x-1\right)-2\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x-2\right)=0\)
<=>\(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(TMDK\right)\\x=2\left(KTMDK\right)\end{cases}}\)
Vậy: x=1
\(\dfrac{5-x}{4x^2-8x}+\dfrac{7}{8x}=\dfrac{x-1}{2x(x-2)}+\dfrac{1}{8x-16}\)
ĐKXĐ: x∉{0;2}
Ta có: \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)
\(\Leftrightarrow\frac{5-x}{4x\left(x-2\right)}+\frac{7}{8x}-\frac{x-1}{2x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}-\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}=0\)
Suy ra: \(10-2x+7x-14-4x+4-x=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;2\right\}\end{matrix}\right.\)
Giai phương trình a) 1/x-1 -3x^2/x^3-1 =2x/x^2+x+1 b) 7/8x+5-x /4x^2 -8x =x-1/2x(x-2) +1/8x-16 c) x+5/x^2-5x -x-5/2x^2 +10x =x+25/2x^2-50 d)|-5x|-|3|=|-16| e) |x-4|=-5 g) |3x-1|=2017 (Mong các bạn giúp đỡ. Cảm ơn)
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