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\(3\left(5x-1\right)-x\left(x-2\right)+x^2-13x=7\)
\(\Leftrightarrow15x-3-x^2+2+x^2-13x=7\)
\(\Leftrightarrow2x-1=7\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=\frac{8}{2}=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-x^2+2x+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow5x^2-20x-41=5x^2-6x+27\)
=>-14x=68
hay x=-34/7
b: \(\Leftrightarrow x^2-25-x^3+6x^2-12x+8-7x^2+x^3+1=\left(x+3\right)^3-x^3-9x^2\)
\(\Leftrightarrow-12x-16=x^3+9x^2+27x+27-x^3-9x^2=27x+27\)
=>-39x=43
hay x=-43/39
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề đúng : Chứng minh : \(\frac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}=\frac{x^2+2x+2}{x-1}\)
Điều kiện : \(x\ne1\)
Phân tích : \(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
\(x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1=x^3+2x-2x^2-\left(x^2-2x+1\right)-1\)
\(=x^3-3x^2+4x-2=\left(x^3-3x^2+3x-1\right)+\left(x-1\right)=\left(x-1\right)^3+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+2\right)\)
Suy ra : \(\frac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}=\frac{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}{\left(x-1\right)\left(x^2-2x+2\right)}=\frac{x^2+2x+2}{x-1}\)
\(\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+\left(x^2-x\right)\)
\(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
\(\Leftrightarrow3x^2+16x-19-3x^2+4x+18=0\)
\(\Leftrightarrow20x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{20}\)
Chúc bạn học tốt !!!