Giải phương trình sau:
\(2x^3-7x^2+7x-\frac{62}{27}=0\\ \)
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29 tháng 7 2019
\(\frac{x^2-4x}{x^2+4x}+\frac{27}{2x^2+7x-4}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4x\right)}{x\left(x+4x\right)}+\frac{27}{2x^2+7x-4}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{x\left(x+4\right)}+\frac{27}{2x^2+8x-x-4}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{x\left(x+4\right)}+\frac{27}{2x\left(x+4\right)-\left(x+4\right)}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x\left(x-4\right)}{x\left(x+4\right)}+\frac{27}{\left(x+4\right)\left(2x-1\right)}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\frac{x-4}{x+4}+\frac{27}{\left(x+4\right)\left(2x-1\right)}=\frac{7-2x}{2x-1}-1\)
\(\Leftrightarrow\left(x-4\right)\left(2x-1\right)+27=\left(7-2x\right)\left(x+4\right)-\left(x+4\right)\left(2x-1\right)\)
\(\Leftrightarrow2x^4-9x+31=-8x+32-4x^2\)
\(\Leftrightarrow2x^2-9x+31+8x-32+4x^2=0\)
\(\Leftrightarrow6x^2-x-1=0\)
\(\Leftrightarrow6x^2+2x-3x-1=0\)
\(\Leftrightarrow2x\left(3x+1\right)-\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\left(\text{nhận}\right)\\x=\frac{1}{2}\left(\text{loại}\right)\end{cases}}\)
\(\Rightarrow x=-\frac{1}{3}\)
Vậy: nghiệm phương trình là \(-\frac{1}{3}\)
BH
20 tháng 2 2019
\(\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1}{\left(x-3\right)\left(2x-1\right)}=\frac{2x+5}{\left(x-3\right)\left(2x-1\right)}\)
\(\frac{\left(x-3\right)\left(x+4\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=\frac{\left(2x+5\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}\)
\(\Rightarrow x^2+x-12+x^2-x-2=2x^2+x-10\Leftrightarrow x=-4\)
20 tháng 2 2019
\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{2x-5}{2x^2-7x+3}-\frac{x+1}{2x^2-7x+3}\)
\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{x+4}{2x^2-7x+3}\)
TH1:\(x+4\ne0\)
\(\Rightarrow2x^2-5x+2=2x^2-7x+3\)
\(\Rightarrow-5x+2=-7x+3\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
TH2:\(x+4=0\)
\(\Rightarrow x=-4\)
3 tháng 2 2019
\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)
\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^2+3x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[2x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(2x+1\right)=0\)
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\(x^3-8x^2-8x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)
......................................................................................
\(2x^3-7x^2+7x-\frac{62}{27}=0\)
\(x^3+\left(x^3-7x^2+7x-\frac{62}{27}\right)=0\)
sau nx lần tra khảo idol
\(x^6-7x^5+7x^4-x^3.\frac{62}{27}=0\)
\(x=x^6-7x^5+7x^4-x^3.\frac{62}{27}\)
HẾT