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4 tháng 2 2017
a) x3+4x2+x-6=0
<=> x3+x2-2x+3x2+3x-6=0
<=>x(x2+x-2)+3(x2+x-2)=0
<=>(x+3)(x2+x-2)=0
<=>(x+3)(x2+2x-x-2)=0
<=>(x+3)[x(x+2)-(x+2)]=0
<=>(x+3)(x-1)(x+2)=0
=> x+3=0 hay
x-1=0 hay
x+2=0
<=> x=-3 hay x=1 hay x=-2
4 tháng 2 2017
b)x3-3x2+4=0
\(\Leftrightarrow x^3-4x^2+4x+x^2-4x+4=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left\{\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
19 tháng 8 2018
Bài 4 : Tìm x biết:
a, 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7)(2x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b, x2 + 36 = 12x
\(\Leftrightarrow\) x2 + 36 - 12x = 0
\(\Leftrightarrow\) x2 - 2.x.6 + 62 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x = 6
19 tháng 8 2018
e, (x - 2)2 - 16 = 0
\(\Leftrightarrow\) (x - 2)2 - 42 = 0
\(\Leftrightarrow\) (x - 2 - 4)(x - 2 + 4) = 0
\(\Leftrightarrow\) (x - 6)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
f, x2 - 5x -14 = 0
\(\Leftrightarrow\) x2 + 2x - 7x -14 = 0
\(\Leftrightarrow\) x(x + 2) - 7(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(x - 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)
5 tháng 11 2017
\(a,x^4-4x^3+x^2-4x=0\)
\(\Rightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Rightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+x\right)=0\)
\(\Rightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
\(b,x^3-5x^2+4x-20=0\)
\(\Rightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Rightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\)
\(\Rightarrow x=5\)
AN
5 tháng 11 2017
a) \(x^4-4x^3+x^2-4x=0\)
\(\Leftrightarrow\left(x^4-4x^3\right)+\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^3+x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x^2+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x^2=-1\left(loai\right)\end{matrix}\right.\)
Vậy x=0; x=4
b) \(x^3-5x^2+4x-20=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(4x-20\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=-4\left(loai\right)\end{matrix}\right.\)
Vậy x=5
27 tháng 5 2022
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
KT
26 tháng 4 2018
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
Suy ra \(X=\frac{\sqrt{3}}{2}\)