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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) \(3y=0\)
\(\Leftrightarrow y=0\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x+3\right)}=0\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2.2x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\frac{x^2+x}{2\left(x-3\right)\left(x+1\right)}+\frac{x^2-3x}{2\left(x-3\right)\left(x+1\right)}-\frac{4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)
=>\(2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
=>\(x=0\)
\(x=3\)
\(\text{GIẢI :}\)
ĐKXĐ : \(x\ne\pm1\)
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}=\frac{x+1}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}-\frac{x+1}{x^2-1}=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow\text{ }2\left(x-1\right)+x\left(x+1\right)-(x+1)=0\)
\(\Leftrightarrow\text{ }2\left(x-1\right)+\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1\text{ (loại)}\\x=-3\text{ (Chọn)}\end{cases}}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-3\right\}\).
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}.\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)\(đk:x\ne\pm1\)
\(< =>\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left[\frac{7}{6}.\frac{6}{7}+\left(1\right)\right]x+1}{x^2-1}\)
\(< =>\frac{2x-2+x^2+x}{x^2+x-x-1}=\frac{2x+1}{x^2-1}\)\(< =>\frac{x^2+3x-2}{x^2-1}=\frac{2x-1}{x^2-1}\)
\(< =>x^2+2x-2=2x-1\)\(< =>x^2+2x-2x-2+1=0\)
\(< =>x^2-1=0< =>x^2=1\)\(< =>x=\pm1\)\(\left(ktmđk\right)\)
Vậy phương trình trên vô nghiệm
\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x+6}ĐKXĐ:x\ne1;2;3;-6\)
\(\frac{\left(x-2\right)\left(x-3\right)\left(x+6\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}+\frac{2.\left(x-1\right)\left(x-3\right)\left(x+6\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)\left(x+6\right)}+\frac{3.\left(x-1\right)\left(x-2\right)\left(x+6\right)}{\left(x-3\right)\left(x-2\right)\left(x-1\right)\left(x+6\right)}=\frac{6.\left(x-1\right)\left(x-3\right)\left(x-2\right)}{\left(x+6\right)\left(x-1\right)\left(x-3\right)\left(x-2\right)}\)
\(14x^2-114x+108=-36x^2+66x-36\)
\(14x^2-114x+108+36x^2-66x+36=0\)
\(50x^2-180x+144=0\)
\(2\left(5x-6\right)\left(5x-12\right)=0\)
\(2\ne0\)=> vô nghiệm
\(5x-6=0\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\)
hoặc
\(5x-12=0\Leftrightarrow5x=12\Leftrightarrow x=\frac{12}{5}\)
Theo ĐKXĐ => tm
Cái chỗ phân tích dài loằng ngoằng kia ko hiểu thì hỏi tớ nha , tớ cx chưa xem lại vì nó hơi dài
\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\).
\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\ne0\right)\)
<=> x=-1
Vậy x=-1