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1. \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
2. \(\left(x-2\right)\left(6x+2\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right).7x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\7x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
3.
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4.
\(x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x_1=0;x_2=2\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow-3x^2-14x+5=0\)
\(\Leftrightarrow\left(-3x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)
\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)