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ĐK: x \(\ne\)-1; x \(\ne\)2
\(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)
<=> \(\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3}{\left(x+1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
<=> x2 - 4 + 3x + 3 = 3 + x2 - x - 2
<=> x2 + 3x - x2 + x = 1 + 1
<=> 4x = 2
<=> x = 1/2
Vậy S = {1/2}
phương trình tương đương với 1+\(\frac{1}{x}+1+\frac{1}{x+3}\)=1+\(\frac{1}{x+1}+1+\frac{1}{x+2}\)\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+3}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\Leftrightarrow\frac{2x+3}{x\left(x+3\right)}=\frac{2x+3}{\left(x+1\right)\left(x+2\right)}\)\(\Leftrightarrow\left(2x+3\right)\left(\frac{1}{x\left(x+3\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\)=0
\(\Leftrightarrow\left(2x+3\right)\left(\frac{\left(x+1\right)\left(x+2\right)-x\left(x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{2}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)\(\Leftrightarrow2x+3=0\Leftrightarrow x=\frac{-3}{2}\)
\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)
Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)
\(\Leftrightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)
Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)
\(b,ĐKXĐ:x\ne\pm1;-3\)
Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)
\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)
\(\Leftrightarrow9x^2+14x+13=0\)
\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)
Pt vô nghiệm
\(c,ĐKXĐ:x\ne1\)
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)
Kết hợp vs ĐKXĐ được x = -1
Vậy pt có nghiệm duy nhất x = -1
làm lần lượt nha(bài nào k bt bỏ qua)
\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow-2x-5=4\)
\(\Rightarrow-2x=9\)
\(\Rightarrow x=\frac{9}{-2}\)
a, \(2+\frac{3}{x-5}=1\Leftrightarrow\frac{3}{x-5}=-1\)
\(\Leftrightarrow x-5=\frac{3}{-1}=-3\Leftrightarrow x=2\)
Vậy .............
b, ....................
\(\Leftrightarrow\frac{x-9}{x^2-3^2}-\frac{2}{x+3}=\frac{1}{x-3}\)
\(\Leftrightarrow\frac{x-9}{\left(x-3\right)\left(x+3\right)}-\frac{2x-6}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x-9-2x+6-x+3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-2x}{\left(x-3\right)\left(x+3\right)}=0\Rightarrow-2x=0\Rightarrow x=0\)
Vậy .............
Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m
Bài 2:
a) \(x+x^2=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(0x-3=0\)
\(\Leftrightarrow0x=3\)
\(\Rightarrow vonghiem\)
c) \(3y=0\)
\(\Leftrightarrow y=0\)
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
ĐKXĐ : x ≠ ±1
pt <=> \(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)
<=> \(\frac{1}{x+1}+\frac{2}{\left(x-1\right)^2\left(x+1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)
<=> \(\frac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}+\frac{2}{\left(x-1\right)^2\left(x+1\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=0\)
<=> \(\frac{x^2-2x+1+2+3x-3}{\left(x-1\right)^2\left(x+1\right)}=0\)
<=> \(\frac{x^2+x}{\left(x-1\right)^2\left(x+1\right)}=0\)
<=> \(\frac{x\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=0\)
<=> \(\frac{x}{\left(x-1\right)^2}=0\)
=> x = 0 ( tm )
Vậy phương trình có nghiệm x = 0