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\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ 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 ~ )
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
#)Giải :
Bài 1 :
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(\Leftrightarrow144x^2+216x+81=144x^2-480x+400\)
\(\Leftrightarrow144x^2+216=144x^2-480x+319\)
\(\Leftrightarrow696x=319\)
\(\Leftrightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Leftrightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x=-1\)
a) 9(4x + 3)2 = 16(3x - 5)2
=> [3(4x + 3)]2 - [4(3x - 5)]2 = 0
=> (12x + 9)2 - (12x - 20)2 = 0
=> (12x + 9 - 12x + 20)(12x + 9 + 12x - 20) = 0
=> 29.(24x - 11) = 0
=> 2x - 11 = 0
=> 2x = 11
=> x = 11 : 2 = 11/2
b) (x3 - x2)2 - 4x2 + 8x - 4 = 0
=> (x3 - x2)2 - (2x - 2)2 = 0
=> (x3 - x2 - 2x + 2)(x3 - x2 + 2x - 2) = 0
=> [x2(x - 1) - 2(x - 1)][x2(x - 1) + 2(x - 1)] = 0
=> (x2 - 2)(x - 1)(x2 + 2)(x - 1) = 0
=> (x2 - 2)(x2 + 2)(x - 1)2 = 0
=> x2 - 2 = 0
hoặc : x2 + 2 = 0
hoặc : (x - 1)2 = 0
=> x2 = 2
hoặc : x2 = -2 (vl)
hoặc : x - 1 = 0
=> \(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
hoặc : x = 1
Vậy ...
c) x5 + x4 + x3 + x2 + x + 1 = 0
=> x4(x +1) + x2(x + 1) + (x + 1) = 0
=> (x4 + x2 + 1)(x + 1) = 0
=> \(\orbr{\begin{cases}x^4+x^2+1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x^4+x^2=-1\left(vl\right)\\x=-1\end{cases}}\) (vì x4 \(\ge\)0 \(\forall\)x; x2 \(\ge\)0 \(\forall\)x => x4 + x2 \(\ge\)0 \(\forall\)x)
=> x = -1
\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(TH2:x+6=0\Leftrightarrow x=-6\)
\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( 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,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
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 \(S=\left\{1;-1;2\right\}\)
(x+1)2=x+1
x2+2x+1-x-1=0
x2+x=0
x(x+1)=0
*x=0
*x+1=0 =>x=-1
(x+1)2=x+1
<=>(x+1)2-(x+1)=0
<=>(x+1)(x+1-1)=0
<=>(x+1).x=0
<=>x+1=0 hoặc x=0
<=>x=-1 hoặc x=0
x2+8x+16=0
<=>(x+4)2=0
<=>x+4=0
<=>x=-4
4.(x-1)2-9.(x+2)2=0
<=>[2.(x-1)]2-[3.(x+2)]2=0
<=>[(2x-2)-(3x+6)][(2x-2)+(3x+6)]=0
<=>(2x-2-3x-6)(2x-2+3x+6)=0
<=>(-x-8)(5x+4)=0
<=>-x-8=0 hoặc 5x+4=0
<=>x=-8 hoặc x=-4/5
(x+3)2=9
<=>(x+3)2-9=0
<=>(x+3-3)(x+3+3)=0
<=>x.(x+6)=0
<=>x=0 hoặc x+6=0
<=>x=0 hoặc x=-6
4x2-12x=-9
<=>4x2-12x+9=0
<=>(2x-3)2=0
<=>2x-3=0
<=>x=3/2
(x-2)2=(3x+1)2
<=>(x-2)2-(3x+1)2=0
<=>[(x-2)-(3x+1)][(x-2)+(3x+1)]=0
<=>(x-2-3x-1)(x-2+3x+1)=0
<=>(-2x-3)(4x-1)=0
<=>-2x-3=0 hoặc 4x-1=0
<=>x=-3/2 hoặc x=1/4
64=(x-3)2
<=>64-(x-3)2=0
<=>[8-(x-3)][8+(x-3)]=0
<=>(8-x+3)(8+x-3)=0
<=>(11-x)(x+5)=0
<=>11-x=0 hoặc x+5=0
<=>x=11 hoặc x=-5