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Bài làm
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow x=-\frac{4}{6}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy x = -2/3 là nghiệm.
@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4
Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
a) ĐKXĐ: \(x\notin\left\{\frac{1}{3};\frac{-11}{3}\right\}\)
Ta có: \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)
\(\Leftrightarrow\frac{2\left(1-3x\right)\left(3x+11\right)}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}=\frac{\left(3x+11\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}-\frac{3\left(1-3x\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}\)
\(\Leftrightarrow-18x^2-60x+22=9x^2+66x+121-3\left(1-6x+9x^2\right)\)
\(\Leftrightarrow-18x^2-60x+22-9x^2-66x-121+3\left(1-6x+9x^2\right)=0\)
\(\Leftrightarrow-27x^2-126x-99+3-18x+27x^2=0\)
\(\Leftrightarrow-144x-96=0\)
\(\Leftrightarrow-144x=96\)
hay \(x=\frac{-2}{3}\)(tm)
Vậy: \(x=\frac{-2}{3}\)
ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)
\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
\(15-20x+6x-12=0\)
\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn
a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)
Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)
\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow9\left(2x-1\right)=0\)
mà 9≠0
nên 2x-1=0
⇔2x=1
hay \(x=\frac{1}{2}\)(tm)
Vậy: \(x=\frac{1}{2}\)
b)ĐKXĐ: x≠0
Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)
\(\Leftrightarrow x+\frac{1}{x}-x^2-\frac{1}{x^2}=0\)
\(\Leftrightarrow\frac{x^3}{x^2}+\frac{x}{x^2}-\frac{x^4}{x^2}-\frac{1}{x^2}=0\)
\(\Leftrightarrow x^3+x-x^4-1=0\)
\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)
\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)(1)
Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)
Từ (1) và (2) suy ra x-1=0
hay x=1(tm)
Vậy: x=1
c) ĐKXĐ: x≠0
Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)
\(\Leftrightarrow\frac{1}{x}+2-\left(\frac{1}{x}+2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(2-x^2-2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\cdot\left(-x^2\right)=0\)(3)
Ta có: 1≠0
x≠0
Do đó: \(\frac{1}{x}\ne0\)
\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)
Từ (3) và (4) suy ra x=0(ktm)
Vậy: x∈∅
d) ĐKXĐ: x≠0
Ta có: \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}+x-1-\frac{1}{x}\right)\left(x+1+\frac{1}{x}-x+1+\frac{1}{x}\right)=0\)
\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)
\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)
mà 4≠0
và x≠0
nên \(1+\frac{1}{x}=0\)
\(\Leftrightarrow\frac{1}{x}=-1\)
hay x=-1(tm)
Vậy: x=-1
ĐKXĐ: \(x\ne\pm2\)
\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\left(tm\right)\)
Vậy: \(S=\left\{-\frac{7}{23}\right\}\)
=.= hk tốt!!
\(\text{ĐKXĐ}\: :\: x\ne\pm2\)
\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
Khử mẫu : \(\left(-6x^2-12x+x+2\right)+\left(9x^2-18x+4x-8\right)=3x^2-2x+1\)
\(\Leftrightarrow-23x=7\Leftrightarrow x=\frac{7}{23}\).