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MN
1 tháng 2 2020
1) \(x^4-2x^2-144x+1295=0\)
\(\Rightarrow\)Cậu xem lại đề thử xem nhé !
2) \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-1\right)-24=0\)
\(\Leftrightarrow x^4+2x^3-x^2-2x-24=0\)
\(\Leftrightarrow x^4+x^3+4x^2+x^3+x^2+4x-6x^2-6x-24=0\)
\(\Leftrightarrow x^2\left(x^2+x+4\right)+x\left(x^2+x+4\right)-6\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x-2x-6\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left(x^2+x+4\right)=0\)
\(\Leftrightarrow\)\(x+3=0\)
hoặc \(x-2=0\)
hoặc \(x^2+x+4=0\)
\(\Leftrightarrow\)\(x=-3\left(tm\right)\)
hoặc \(x=2\left(tm\right)\)
hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)
3) \(x^4-2x^3+4x^2-3x-10=0\)
\(\Leftrightarrow x^4+x^3-3x^3-3x^2+7x^2+7x-10x-10=0\)
\(\Leftrightarrow x^3\left(x+1\right)-3x^2\left(x+1\right)+7x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-3x^2+7x-10\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3-2x^2-x^2+2x+5x-10\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x-2\right)-x\left(x-2\right)+5\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)=0\)
\(\Leftrightarrow\)\(x+1=0\)
hoặc \(x-2=0\)
hoặc \(x^2-x+5=0\)
\(\Leftrightarrow x=-1\left(tm\right)\)
hoặc \(x=2\left(tm\right)\)
hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là :\(S=\left\{-1;2\right\}\)
DT
2
2 tháng 2 2019
\(3x^3+2x^2+2x+3=0\)
\(\Leftrightarrow3\left(x^3+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2-x+3\right)=0\)
Mà \(3x^2-x+3=3\left[\left(x-\frac{1}{6}\right)^2+\frac{35}{36}\right]>0\forall x\)
Do đó: \(x+1=0\Leftrightarrow x=-1\)
Tập nghiệm: \(S=\left\{-1\right\}\)
2 tháng 2 2019
\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
\(\Leftrightarrow\left[\left(x-1\right)+\left(2x+3\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=27x^3+8\)
\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9\right)=27x^3+8\)
\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(6x^2-15x-9\right)=0\)(Chuyển vế)
\(\Leftrightarrow3\left(3x+2\right)\left(2x^2-5x-3\right)=0\)
\(\Leftrightarrow3\left(3x+2\right)\left(x-3\right)\left(2x+1\right)=0\)
Tập nghiệm: \(S=\left\{-\frac{2}{3};3;-\frac{1}{2}\right\}\)
14 tháng 10 2018
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
HC
0
TM
21 tháng 7 2016
1/\(x^2+5x+6=0\)
=>\(x^2+2x+3x+6=0\)
=>\(x\left(x+2\right)+3\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)
Các câu sau làm tương tự câu 1, tách ghép khéo léo sẽ ra :)
\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=0\)
\(4x^2-4x+1-\left(4x^2-12x+x-3\right)=0\)
\(4x^2-4x+1-\left(4x^2-11x-3\right)=0\)
\(4x^2-4x+1-4x^2+11x+3=0\)
\(7x+4=0\)
\(7x=-4\)
\(x=\frac{-4}{7}\)
Vậy \(x=\frac{-4}{7}\)
\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-\left(4x^2-12x+x-3\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+12x-x+3=0\)
\(\Leftrightarrow\left(4x^2-4x^2\right)+\left(-4x+12x-x\right)+\left(1+3\right)=0\)
\(\Leftrightarrow7x+4=0\)
\(\Leftrightarrow x=-\frac{4}{7}\)