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NN
8 tháng 8 2023
a/
\(x^3-4x^2-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
b/
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
c/
\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^4\left(x-1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^4-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)
NT
3
23 tháng 8 2021
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
19 tháng 11 2021
a. x( x+ 3)= 0
⇔ x= 0 hoặc x+ 3= 0
⇔ x= 0 x = -3
b. x( 2x− 1)+ 2( 2x− 1) =0
⇔ ( 2x− 1)(x+ 2) =0
⇔ 2x− 1 =0 hoặc x+ 2 =0
⇔ 2x =1 x = -2
⇔ x =\(\dfrac{1}{2}\) x = -2
28 tháng 12 2021
\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
4 tháng 8 2018
\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)