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Theo bài ra ,ta có :
\(\frac{x+1}{x-2}-\frac{1}{x}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{x+1}{x-2}-\frac{1}{x}=\frac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\left(ĐKXĐ:x\ne0;x\ne2;x\ne-2\right)\)
Quy đồng và khử mẫu ta được
\(x\left(x+1\right)\left(x+2\right)-\left(x-2\right)\left(x+2\right)=2x\left(x^2+2\right)\)
\(\Leftrightarrow\left(x^2+x\right)\left(x+2\right)-\left(x-2\right)\left(x+2\right)=2x^3+4x\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-x+2\right)=2x^3+4x\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+2\right)=2x^3+4x\)
\(\Leftrightarrow x^3+2x+2x^2+4=2x^3+4x\)
\(\Leftrightarrow x^3-2x^3+2x^2+2x-4x+4=0\)
\(\Leftrightarrow-x^3+2x^2-2x+4=0\)
\(\Leftrightarrow-\left(x^3-2x^2+2x-4\right)=0\)
\(\Leftrightarrow-\left(x^2\left(x-2\right)+2\left(x-2\right)\right)=0\)
\(\Leftrightarrow-\left(\left(x-2\right)\left(x^2+2\right)\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow2-x=0\)( Vì x2 + 2 luôn luôn > 2 với mọi x )
\(\Leftrightarrow x=2\)(Không TMĐKXĐ) ( Loại )
Vậy S={rỗng}
Chúc bạn học tốt =))
Bạn chú ý cách viết phương trình.
Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.
\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=4\left(x+\frac{1}{x}\right)^2\left(2-x^2-\frac{1}{x^2}\right)+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4\left(x+\frac{1}{x}\right)^2\left(x-\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4\left(x^2-\frac{1}{x^2}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)
\(=-4x^4+8-\frac{4}{x^4}+4x^4+8+\frac{4}{x^4}\)
\(=16\)
Phương trình đã cho trở thành
\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
\(\Rightarrow x\left(x+3\right)=10=2.\left(2+3\right)\)
\(\Rightarrow x=2\)
pt <=> \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
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
\(x-1=0\)
\(x=1\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\)
\(\Leftrightarrow x^2+8x+16=16\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow x\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)
V...\(S=\left\{-8\right\}\)
^^
bạn ghi sai đề ở chỗ \(\left(x+\frac{1}{x}\right)^2\)chứ ko phải \(\left(x+\frac{1}{x^2}\right)^2\)nhé
\(\frac{x-1}{x^2-x+1}-\frac{x+1}{x^2+x+1}=\frac{10}{x\left(x^4+x+1\right)}\)
\(\Leftrightarrow\frac{x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2+x+1\right)-10}{x\left(x^4+x^2+1\right)}=0\)
\(\Rightarrow x\left(x^3-1\right)-x\left(x^3+1\right)-10=0\)
\(\Leftrightarrow x^4-x-x^4-x-10=0\)
\(\Leftrightarrow-2x-10=0\)
\(\Leftrightarrow x=-5\)