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1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(TH2:x+6=0\Leftrightarrow x=-6\)
\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( ns vô nghiệm cx ko sai nha )
\(\left(2x+5\right)^2=\left(3x-1\right)^2\)
\(2x+5=3x-1\)
\(2x-3x=-1-5\)
\(-1x=-6\)
\(x=6\)
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{2}{3};-1;\dfrac{1}{2}\right\}\)
\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x^2\right)=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)\left(1+x\right)-\left(1-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1-x-x-3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(-3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\-3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-1\right\}\)
\(c,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\-5x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{7}{5}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-2;\dfrac{7}{5}\right\}\)
\(d,x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
Vậy phương trình có nghiệm duy nhất x = -1
\(e,x^3-7x+6=0\)
\(\Leftrightarrow x^3-4x-3x+6=0\)
\(\Leftrightarrow x\left(x^2-4\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;2;-3\right\}\)
\(f,x^4-4x^3+12x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)-\left(4x^3-12x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)-4x\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x^2-3-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3>0\forall x\\x^2-4x-3>0\forall x\end{matrix}\right.\)
Vậy phương trình vô nghiệm
\(g,x^5-5x^3+4x=0\)
\(\Leftrightarrow x\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^4-4x^2-x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^2-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\) hoặc x = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\\x=-1\end{matrix}\right.\) hoặc x =0
Vậy tập nghiệm của pt \(S=\left\{0;1;-1;2;-2\right\}\)
\(h,x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^4-x^2\right)-\left(4x^3-4x\right)+\left(4x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4x\left(x^2-1\right)+4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{1;-1;2\right\}\)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
\(x^2-2x=24\)
<=> \(x^2-2x-24=0\)
<=> \( \left(x+4\right)\left(x-6\right)=0\)
<=> \(\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
Vậy....
\(a,\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow\left(x+2\right)^2+4-x^2=0\)
\(\Leftrightarrow\left(2+x\right)^2+\left(2-x\right)\left(2+x\right)=0\)
\(\Leftrightarrow\left(2+x\right)\left(2+x+2-x\right)=0\)
\(\Leftrightarrow4\left(2+x\right)=0\)
\(\Leftrightarrow2+x=0\)
\(\Leftrightarrow x=-2\)
\(c,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)
\(\Leftrightarrow2x+255=0\)
\(\Leftrightarrow x=-127,5\)
Lời giải:
1.
$x^3-x(x^2+2)=6$
$\Leftrightarrow -2x=6\Leftrightarrow x=-3$
2.
\((x-2)^2-3(x-2)=0\Leftrightarrow (x-2)(x-2-3)=0\)
\(\Leftrightarrow (x-2)(x-5)=0\Rightarrow \left[\begin{matrix} x=2\\ x=5\end{matrix}\right.\)
3.
\(x^2+2x+1=0\Leftrightarrow (x+1)^2=0\Rightarrow x=-1\)
4.
\(4x^2-12x=-9\Leftrightarrow 4x^2-12x+9=0\)
\(\Leftrightarrow (2x)^2-2.2x.3+3^2=0\Leftrightarrow (2x-3)^2=0\)
\(\Rightarrow x=\frac{3}{2}\)