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Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)
\(< =>\frac{-2}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)
\(< =>-x^2+5x-6-x^2-x+5=0\)
\(< =>-2x^2+4x-1=0\)
\(< =>2x^2-4x+1=0\)
đến đây thì pt bậc 2 dể rồi
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)
\(< =>\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x^2-1\right)\left(x-1\right)}=-\frac{3}{x^2-1}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x+1\right)\left(x-1\right)^2}=\frac{-3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)
\(< =>2+3x-3+x^2-2x+1=0\)
\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)
Ta có: \(\dfrac{2x}{x^2-x+1}-\dfrac{x}{x^2+x+1}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x\left(x^2+x+1\right)-x\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x^3+2x^2+2x-x^3+x^2-x}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{x^3+3x^2+x}{\left(x^2+1\right)^2-x^2}=\dfrac{5}{3}\)
\(\Leftrightarrow3x^3+9x^2+3x=5\left(x^4+2x^2+1-x^2\right)\)
\(\Leftrightarrow3x^3+9x^2+3x=5x^4+5x^2+5\)
\(\Leftrightarrow5x^4+5x^2+5-3x^3-9x^2-3x=0\)
\(\Leftrightarrow5x^4-3x^3-4x^2-3x+5=0\)
\(\Leftrightarrow5x^4-5x^3+2x^3-2x^2-2x^2+2x-5x+5=0\)
\(\Leftrightarrow5x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3+2x^2-2x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3-5x^2+7x^2-7x+5x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5x^2\left(x-1\right)+7x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(5x^2+7x+5\right)=0\)
mà \(5x^2+7x+5>0\forall x\)
nên x-1=0
hay x=1
=4x^2-4x+1+x^3-27-4(x^2-16)
=4x^2-4x+1+x^3-27-4x^2+64
=x^3-4x+38
\(2\left(x^2-x\right)-x\left(x+2\right)+4=0\)
\(\Leftrightarrow2x^2-2x-x^2-2x+4=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
⇔ ( x + 2 )( x - 1 ) = 0 ⇔
Vậy phương trình có tập nghiệm là S = { - 2;1 }.
1. Đặt $x^2+x=a$ thì pt trở thành:
$a^2+4a=12$
$\Leftrightarrow a^2+4a-12=0$
$\Leftrightarrow (a-2)(a+6)=0$
$\Leftrightarrow a-2=0$ hoặc $x+6=0$
$\Leftrightarrow x^2+x-2=0$ hoặc $x^2+x+6=0$
Dễ thấy $x^2+x+6=0$ vô nghiệm.
$\Rightarrow x^2+x-2=0$
$\Leftrightarrow (x-1)(x+2)=0$
$\Leftrightarrow x=1$ hoặc $x=-2$
2.
$x(x-1)(x+1)(x+2)=24$
$\Leftrightarrow [x(x+1)][(x-1)(x+2)]=24$
$\Leftrightarrow (x^2+x)(x^2+x-2)=24$
$\Leftrightarrow a(a-2)=24$ (đặt $x^2+x=a$)
$\Leftrightarrow a^2-2a-24=0$
$\Leftrightarrow (a+4)(a-6)=0$
$\Leftrightarrow a+4=0$ hoặc $a-6=0$
$\Leftrightarrow x^2+x+4=0$ hoặc $x^2+x-6=0$
Nếu $x^2+x+4=0$
$\Leftrightarrow (x+\frac{1}{2})^2=\frac{1}{4}-4<0$ (vô lý - loại)
Nếu $x^2+x-6=0$
$\Leftrightarrow (x-2)(x+3)=0$
$\Leftrightarrow x-2=0$ hoặc $x+3=0$
$\Leftrightarrow x=2$ hoặc $x=-3$
\(\left(x^2+x\right)-4\left(x^2+x\right)=12\)
\(\Leftrightarrow x^2+x-4x^2-4x-12=0\)
\(\Leftrightarrow-3x^2-3x-12=0\)
\(\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow\left(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) (vô lí)
-Vậy S=∅