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giải phương trình có chứa ẩn ở mẫu
Answer:
e) \(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\left(ĐK:x\ne2;x\ne4\right)\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)
\(\Leftrightarrow x^2-7x+12-x^2+4x-4=\frac{16}{5}.\left(x-2\right)\left(x-4\right)\)
\(\Leftrightarrow-3x+8=\frac{16}{5}.\left(x^2-6x+8\right)\)
\(\Leftrightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)
\(\Leftrightarrow-\frac{16}{5}x^2+\frac{81}{5}x-\frac{88}{5}=0\)
\(\Leftrightarrow-\frac{16}{5}.\left(x^2-\frac{81}{16}x+\frac{11}{2}\right)=0\)
\(\Leftrightarrow x^2-\frac{81}{16}x+\frac{6561}{1024}-\frac{929}{1024}=0\)
\(\Leftrightarrow\left(x-\frac{81}{32}\right)^2=\frac{929}{1024}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{81}{32}=\frac{\sqrt{929}}{32}\\x-\frac{81}{32}=-\frac{\sqrt{929}}{32}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}}\)
f) \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\left(ĐK:x\ne2;x\ne4\right)\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Leftrightarrow x^2-4x-3x+12+x^2-4x+4=-x^2+4x+2x-8\)
\(\Leftrightarrow x^2+x^2+x^2-4x-3x-4x-4x-2x+12+4+8=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\3x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{8}{3}\end{cases}}}\)
Answer:
e) \(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\left(ĐK:x\ne2;x\ne4\right)\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)
\(\Leftrightarrow x^2-7x+12-x^2+4x-4=\frac{16}{5}.\left(x-2\right)\left(x-4\right)\)
\(\Leftrightarrow-3x+8=\frac{16}{5}.\left(x^2-6x+8\right)\)
\(\Leftrightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)
\(\Leftrightarrow-\frac{16}{5}x^2+\frac{81}{5}x-\frac{88}{5}=0\)
\(\Leftrightarrow-\frac{16}{5}.\left(x^2-\frac{81}{16}x+\frac{11}{2}\right)=0\)
\(\Leftrightarrow x^2-\frac{81}{16}x+\frac{6561}{1024}-\frac{929}{1024}=0\)
\(\Leftrightarrow\left(x-\frac{81}{32}\right)^2=\frac{929}{1024}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{81}{32}=\frac{\sqrt{929}}{32}\\x-\frac{81}{32}=-\frac{\sqrt{929}}{32}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}}\)
f) \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\left(ĐK:x\ne2;x\ne4\right)\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Leftrightarrow x^2-4x-3x+12+x^2-4x+4=-x^2+4x+2x-8\)
\(\Leftrightarrow x^2+x^2+x^2-4x-3x-4x-4x-2x+12+4+8=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\3x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{8}{3}\end{cases}}}\)