trong quá trình bạn xem bài mk thấy chỗ nào sai dấu thì sửa giùm mk nha trong quá trình làm mk cx có thể sai sót nhầm lẫn nha
\( m){x^4} + 2{x^3} - 13{x^2} - 14x + 24 = 0\\ \Leftrightarrow {x^4} - {x^3} + 3{x^3} - 3{x^2} - 10{x^2} + 10x - 24x + 24 = 0\\ \Leftrightarrow {x^3}\left( {x - 1} \right) + 3{x^2}\left( {x - 1} \right) - 10x\left( {x - 1} \right) - 24\left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 3{x^2} - 10x - 24} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^3} + 2{x^2} + {x^2} + 2x - 12x - 24} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left[ {{x^2}\left( {x + 2} \right) + x\left( {x + 2} \right) - 12\left( {x + 2} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + x - 12} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {{x^2} + 4x - 3x - 12} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left[ {x\left( {x + 4} \right) - 3\left( {x + 4} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x - 3} \right)\left( {x + 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 1 = 0\\ x + 2 = 0\\ x - 3 = 0\\ x + 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 2\\ x = 3\\ x = - 4 \end{array} \right. \)
\(n)\dfrac{x+1}{99}+\dfrac{x+2}{98}=\dfrac{x+3}{97}+\dfrac{x+4}{96}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right) = \left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}=\dfrac{x+100}{97}+\dfrac{x+100}{96}\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}-\dfrac{x+100}{97}-\dfrac{x+100}{96}=0\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\right)=0 \)
Mà \(\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{97}-\dfrac{1}{96}\ne0 \\ \)
\(\Rightarrow x+100=0\\ \Rightarrow x=-100\\ \)
Vậy \(x=-100\)
\(i)\) Ta có: \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)
\(\Leftrightarrow\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)=1680\)
\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)=1680\)
Gọi: \(x^2-11x+29=a\)
\(\Rightarrow\left(a-1\right)\left(a+1\right)=1680\)
\(\Leftrightarrow a^2-1=1680\)
\(\Leftrightarrow a^2=1681\)
\(\Leftrightarrow a=\pm41\)
* Nếu \(a=-41\)
\(\Leftrightarrow x^2-11x+29=-41\)
\(\Leftrightarrow x^2-11x+70=0\)
\(\Leftrightarrow x^2-2.\dfrac{11}{2}x+\dfrac{121}{4}-\dfrac{121}{4}+70=0\)
\(\Leftrightarrow\left(x-\dfrac{11}{2}\right)^2+\dfrac{159}{4}=0\) ( vô nghiệm )
*Nếu \(a=41\)
\(\Leftrightarrow x^2-11x+29=41\)
\(\Leftrightarrow x^2-11x-12=0\)
\(\Leftrightarrow x^2+x-12x-12=0\)
\(\Leftrightarrow x\left(x+1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=12\end{matrix}\right.\)
Vây: Tập nghiệm của phương trình là: \(S=\left\{-1;12\right\}\)
\( k){x^3} - {x^2} - 17x - 15 = 0\\ \Leftrightarrow {x^3} + {x^2} - 2{x^2} - 2x - 15x - 15 = 0\\ \Leftrightarrow {x^2}\left( {x + 1} \right) - 2x\left( {x + 1} \right) - 15\left( {x + 1} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} - 2x - 15} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 3x - 5x - 15} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left[ {x\left( {x + 3} \right) - 5\left( {x + 3} \right)} \right] = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ x + 3 = 0\\ x - 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 3\\ x = 5 \end{array} \right. \)
Vậy \(S=\left\{-1;-3;5\right\}\)
\( l){x^3} + 4{x^2} + x - 6 = 0\\ \Leftrightarrow {x^3} - {x^2} + 5{x^2} - 5x + 6x - 6 = 0\\ \Leftrightarrow {x^2}\left( {x - 1} \right) + 5x\left( {x - 1} \right) + 6\left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 5x + 6} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 3x + 2x + 6} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left[ {x\left( {x + 3} \right) + 2\left( {x + 3} \right)} \right] = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 1 = 0\\ x + 3 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - 3\\ x = - 2 \end{array} \right. \)